Brilliant Integration Contest - Season 2 (Part-1)

@Aditya Kumar – Another solution: Let \(\sqrt{(1-x^2)}=t\)

\(\frac{-x}{\sqrt{(1-x^2)}}dx=dt\)

So integration becomes

\(- \int_1^0 ln\sqrt{(1-t^2)}dt\)

\(-\frac{1}{2} \int_1^0 ln(1-t^2)dt\)

\(\frac{1}{2} \int_0^1 ln(1-t^2)dt\)

\(\frac{1}{2} \int_0^1 (ln(1-t)+ln(1+t))dt\)

This can be easily evaluated and comes out to be \(ln2-1\) – Gautam Sharma · 1 year, 10 months ago

@Aditya Kumar – \(\Large \text{Solution to Problem 1:}\)

\[\int_{0}^{1} \dfrac{x \ln(x)}{\sqrt{1-x^2}} dx\] Take \(t = \arcsin(x) \iff dt = \dfrac{dx}{\sqrt{1-x^2}}\) and the limits changes to \(t=0\) to \(t=\dfrac{\pi}{2}\). \[\int_{0}^{\dfrac{\pi}{2}} \sin (t) \ln(\sin(t)) dt\]

Consider,

\[B(x,y) = \int_{0}^{1} t^{x-1} (1-t)^{y-1} dt\]

Taking the transformation \(t=\sin^2 (z) \iff dt = 2\sin(z) \cos (z)dz\), the integral transforms into

\[B(x,y) = 2 \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) dz\]

But \(B(x,y) = \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)}\)

So, \[\int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) dz = \dfrac{1}{2} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)}\]

Differentiate it w.r.t \(x\)

\[2 \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) \ln(\sin(z)) dz = \dfrac{1}{2} \dfrac{\partial}{\partial x} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)} \\ \int_{0}^{\pi /2} \sin ^{2x-1} (z) \cos^ {2y-1} (z) \ln(\sin(z)) dz = \dfrac{1}{4} \dfrac{\Gamma (x) \Gamma(y) }{\Gamma (x+y)} (\psi (x) - \psi(x+y)) \]

Take \(x= 1\) and \(y= \dfrac{1}{2}\).

\[\int_{0}^{\pi / 2} \sin(z) \ln (\sin (z)) dz = \dfrac{1}{4} \dfrac{\Gamma (1) \Gamma(1/2) }{\Gamma (3/2)} (\psi (1) - \psi(3/2)) = \ln (2) -1\] – Surya Prakash · 1 year, 10 months ago

@Aditya Kumar – Solution to Problem 23:

Substitute: \(x\longrightarrow \frac { 1-x }{ 1+x } \)

\[I=\int_0^1 \ln(1+x)\ln(1-x^3)dx=2\int_0^1 \ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right)\frac{dx}{(1+x)^2}\]

On separating the integrands,

\(\ln\left(\frac{2 x (3+x^2)}{(1+x)^3}\right)\ln\left(\frac{2}{1+x}\right) \\\small =\ln^22-\ln2\ln x+\ln2\ln(3+x^2)-4\ln2\ln(1+x)-\ln x\ln(1+x)+3\ln^2(1+x)-\ln(1+x)\ln(3+x^2)\)

1st integral:

\[\int_0^1 \frac{dx}{(1+x)^2}=-\frac1{1+x}\Bigg{|}_0^1=\frac12\]

2nd integral:

\[\int_0^1 \frac{\ln x}{(1+x)^2}dx=\sum_{n\geq1} \frac{(-1)^n}{n}=-\ln2\]

3rd integral:

\[\int_0^1 \frac{\ln(1+x)}{(1+x)^2}dx=-\frac{\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac{dx}{(1+x)^2}dx=\frac12-\frac{\ln2}{2}\]

4th integral:

\[\int_0^1 \frac{\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2x}{(1+x)(3+x^2)}dx \\=-\ln2+\ln3+2\operatorname{Re} \int_0^1 \frac{dx}{(1+x)(x+i\sqrt{3})}=-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}\]

5th integral:

\[\int_0^1 \frac{\ln x \ln(1+x)}{(1+x)^2}dx=-\frac{\ln x\ln(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1\frac1{1+x}\left(\frac{\ln(1+x)}{x}+\frac{\ln x}{1+x}\right)dx \\=-\ln2+\int_0^1 \frac{\ln(1+x)}{x}dx-\int_0^1\frac{\ln(1+x)}{1+x}dx=\frac{\pi^2}{12}-\frac12\ln^22-\ln2\]

6th integral:

\[\int_0^1 \frac{\ln^2(1+x)}{(1+x)^2}dx=-\frac{\ln^2(1+x)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac{2\ln(1+x)}{(1+x)^2}dx \\=1-\ln2-\frac12\ln^22\]

7th integral:

\[\int_0^1 \frac{\ln(1+x)\ln(3+x^2)}{(1+x)^2}dx=-\frac{\ln(1+x)\ln(3+x^2)}{1+x}\Bigg{|}_0^1+\int_0^1 \frac1{1+x}\left(\frac{2x\ln(1+x)}{3+x^2}+\frac{\ln(3+x^2)}{1+x}\right)dx \\=-\ln^22-\ln2+\frac34\ln3+\frac{\pi}{4\sqrt{3}}+2J\]

where,

\[J=\int_0^1\frac{x\ln(1+x)}{(1+x)(3+x^2)}dx=\operatorname{Re} \int_0^1 \frac{\ln(1+x)}{(1+x)(x+i\sqrt{3})}dx \\=-\frac18\ln^22-\operatorname{Re} \,\frac1{i\sqrt{3}-1}\int_0^1 \frac{\ln(1+x)}{x+i\sqrt{3}}dx\]

Now on using the formula:

\[\int_0^1 \frac{\ln(1+x)}{x+a}dx=\ln2 \ln\frac{a+1}{a-1}+\operatorname{Li}_2\left(\frac2{1-a}\right)-\operatorname{Li}_2\left(\frac1{1-a}\right)\]

Putting \(a=i\sqrt{3}\) then gives, after taking the real part, and assuming the principle value of the logarithm,

\[\small J=-\frac18\ln^22+\frac{\pi\sqrt{3}}{12}\ln2+\frac14\operatorname{Re}\text{Li}_2(e^{i\pi/3})-\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(e^{i\pi/3})-\frac14\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})+\frac{\sqrt{3}}{4}\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})\]

Using:

\[\displaystyle \,\, \operatorname{Re}\text{Li}_2(e^{i\pi/3})=\frac{\pi^2}{36}\],

\[\displaystyle \,\, \operatorname{Im}\text{Li}_2(e^{i\pi/3})=\frac1{2\sqrt{3}}\psi_1\left(\frac13\right)-\frac{\pi^2}{3\sqrt{3}}\],

\[\displaystyle \,\,\text{Li}_2(\frac12 e^{i\pi/3})=-\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)-\frac12\ln^2\left(\frac34-i\frac{\sqrt{3}}{4}\right)\],

and

\[\displaystyle \,\, \operatorname{Re}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right)=\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)^2\, 3^n}=\frac14\text{Li}_2\left(-\frac13\right)\]

we have \[\operatorname{Re}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi^2}{72}-\frac18\ln^2\left(\frac34\right)-\frac14\text{Li}_2\left(-\frac13\right)\]

and \[\operatorname{Im}\text{Li}_2(\frac12 e^{i\pi/3})=\frac{\pi}{12}\ln\left(\frac34\right)-\operatorname{Im}\text{Li}_2\left(-\frac{i}{\sqrt{3}}\right).\]

Finally, putting everything together gives: \[I=\ln^22 - \frac18\ln^23 + 2\ln2\ln3 - \frac32\ln3 - 6\ln2 + 6-\frac{\pi}{4\sqrt{3}}(2 + \ln3) - \frac{37\pi^2}{72} + \frac12\psi_1\left(\frac13 \right ) - \frac14\text{Li}_2\left( -\frac13 \right ) + \sqrt{3}\Im\text{Li}_2\left( -\frac{i}{\sqrt{3}} \right )\]

\[...\]

This was not original. I found it very awesome, so I thought to share it here. – Aditya Kumar · 1 year, 9 months ago

@Surya Prakash – \(\large{Solution~to~Problem~ 2}\)

Firstly I would add a proof of my solution

\(\large{\displaystyle \int _{0}^{\infty} e^{-ax}x^{n-1} dx=\frac{\Gamma (n)}{a^{n}}}\) where \(\Gamma (n)\) is Gamma function

Replace \(a \rightarrow a+ib\) where \(i =\sqrt{-1}\)

\(\large{\displaystyle \int _{0}^{\infty} e^{-ax}e^{-ibx}x^{n-1} dx=\frac{\Gamma (n)}{(a+ib)^{n}}}\)

Put \(a=r \cos y ~ and ~ b=r \sin y\).

So that \(r^2=a^2+b^2\) and \(y=\arctan \left(\frac{b}{a}\right)\)

Using de moviers theorem

\(\large{\displaystyle \int _{0}^{\infty} e^{-ax}(\cos bx-i \sin bx)x^{n-1} dx=\frac{\Gamma (n)}{r^{n}}(\cos ny+\sin ny)^{-1}}\)

Comparing the real and imaginary parts we finally get

\(\large{\displaystyle \int _{0}^{\infty}x^{n-1} e^{-ax}(\cos bx) dx=\frac{\Gamma (n)}{r^{n}}(\cos ny)}\)

now in the given integral put \(x^{3}=t\)

The integral now becomes

\(\large{\frac{1}{3} \displaystyle \int^{\infty}_{0} t^{-\frac{1}{3}}e^{-t} \cos (t) dt}\)

here \(n=\frac{2}{3}\);\(a=1\) and \(b=1\)

Thus we get \(\large{\frac{1}{3}\frac{\Gamma \left(\frac{2}{3}\right) \cos \left(\frac{\pi}{6}\right)}{2^{\frac{1}{3}}}}\)

\(\large{\boxed{\frac{\sqrt{3} \Gamma \left(\frac{2}{3}\right)}{3.2^{\frac{4}{3}}}}}\) – Tanishq Varshney · 1 year, 10 months ago

@Pi Han Goh – \(\text{Solution to Problem 24}\)

\(\large{\displaystyle \int ^{\infty}_{0} \frac{x^{m-1}}{1+x}dx=\frac{\pi}{\sin m \pi} \quad \quad 0<m<1}\)

Refer to problem 18 for this step.

Now in the integral \(\tan x \rightarrow t\)

\(\large{\displaystyle \int ^{\infty}_{0} \frac{(1+t^2)^3}{1+t^{8}}dt}\)

\(\large{t^8 \rightarrow z}\)

\(\large{\frac{1}{8} \left(\displaystyle \int ^{\infty}_{0} \frac{z^{\frac{7}{8}-1}}{1+z}+ \frac{3z^{\frac{5}{8}-1}}{1+z}+ \frac{3z^{\frac{3}{8}-1}}{1+z}+ \frac{z^{\frac{1}{8}-1}}{1+z} dz \right)}\)

\(\large{\rightarrow \frac{\pi}{4} \left(\frac{1}{\sin \left(\frac{\pi}{8} \right)}+\frac{3}{\cos \left(\frac{\pi}{8} \right)} \right)}\)

\(\large{=\frac{\sqrt{10-\sqrt{2}}}{2} \pi}\) – Tanishq Varshney · 1 year, 9 months ago

@Pi Han Goh – I'll post the outline of the solution as it is a bit cumbersome. Let
\( \displaystyle I(a) = \int_0^{\frac{\pi}{2}} \arctan (a \cos^2 x) dx \Rightarrow I'(a) = \frac{\cos^2 x}{1 + a^2 \cos^4 x } dx = \frac{\sec^2 x}{\sec^4 x + a^2 } dx \) Substitute \( \tan x = t \), to get \( \displaystyle I'(a) = \int_0^{\infty} \frac{1}{t^4 + 2t^2 + (a^2 + 1)} dt \) . Put \(a^2 + 1 = k^2 \) and with some simple rearrangement, we get,
\(\displaystyle 2k I'(a) = \int_0^{\infty} \frac{ 1+ \frac{k}{t^2}}{\left( t - \frac{k}{t} \right)^2 + 2(k+1)} - \int_0^{\infty} \frac{ 1- \frac{k}{t^2}}{\left( t + \frac{k}{t} \right)^2 + 2(-k+1)} \) Substitute \( \displaystyle t - \frac{k}{t} = z \) and \( \displaystyle t + \frac{k}{t} = y \) in the respective integrals and evaluate them easily (They come out to be \( \arctan \) of something ). After all this we can write,
\(\displaystyle I(a) = \frac{\pi}{2\sqrt{2}} \int \frac{1}{\sqrt{a^2 + 1}} \frac{1}{\sqrt{\sqrt{a^2 + 1} +1}} da \)
This can be rearranged to be written as
\( \displaystyle I(a) = \frac{\pi}{2\sqrt{2}} \int \frac{1}{1+ \frac{\sqrt{a^2 + 1} - 1 }{2}} \times \frac{1}{2\sqrt{ \frac{\sqrt{a^2 + 1} - 1 }{2} }} \times \frac{1}{2} \times \frac{a}{\sqrt{a^2 +1}} da \)

This evaluates to \( \displaystyle \frac{\pi}{2} \arctan \left( \sqrt{ \frac{\sqrt{a^2+1} - 1}{2}} \right) \)

The constant of integration is \( 0\) from \( I(0) = 0 \). Put \( a =1729 \) to get the answer. – Sudeep Salgia · 1 year, 10 months ago

@Pi Han Goh – By using integration by parts,

\[\int x \ln \left(\dfrac{1+x}{1-x} \right) dx = \dfrac{x^2}{2} \ln \left(\dfrac{1+x}{1-x} \right) + x - \dfrac{1}{2} \ln \left(1+x \right) + \dfrac{1}{2} \ln \left(1-x \right) \]

So,

\[\begin{align*} \int_{0}^{1} x \ln \left(\dfrac{1+x}{1-x} \right) dx &= \lim_{a \rightarrow 0} \int_{0}^{1-a} x \ln \left(\dfrac{1+x}{1-x} \right) dx \\ &= \lim_{a \rightarrow 0} \dfrac{(1-a)^2}{2} \ln \left(\dfrac{2-a}{a} \right) + 1 - a - \dfrac{1}{2} \ln \left(2-a \right) + \dfrac{1}{2} \ln a \\ &= 1\end{align*}\] – Surya Prakash · 1 year, 10 months ago

@Tanishq Varshney – \[\begin{align*} x^2 + 20 &= (x \sin(x) + 5 \cos(x))^2 + (5 \sin (x) - x \cos(x))^2 - 5 \\ &=(x \sin(x) + 5 \cos(x))^2 + (5 \sin (x) - x \cos(x))^2 - 5 \cos^2(x) - 5\sin^2 (x) -x \cos(x)\sin(x) + x\cos(x) \sin(x) \\ &= (x \sin(x) + 5 \cos(x)) ( 5 \cos(x) + x \sin(x) - \cos(x)) + (5 \sin(x) - x \cos(x))(5 \sin(x) - x \cos(x) - \sin(x)) \\ &= (x \sin(x) + 5 \cos(x)) ( 5 \cos(x) + x \sin(x) - \cos(x)) - (5 \sin(x) - x \cos(x))( x \cos(x) + \sin(x) - 5 \sin (x)) \\ &= (x \sin(x) + 5 \cos(x)) \dfrac{d(5 \sin(x) - x \cos(x))}{dx} - (5 \sin(x) - x \cos(x)) \dfrac{d(x \sin(x) + 5 \cos(x))}{dx} \end{align*}\]

Let \(u=x \sin(x) + 5 \cos(x)\) and \(v=5 \sin(x) - x \cos(x)\). So, \(x^2 + 20 = u\dfrac{dv}{dx} - v\dfrac{du}{dx} = u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}\).

So, \[\int \dfrac{x^2 + 20}{(x \sin(x) + 5 \cos(x))^2} dx = \int \dfrac{ u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}}{u^2} dx= \dfrac{v}{u} = \dfrac{5 \sin(x) - x \cos(x)}{x \sin(x) + 5 \cos(x)}\] – Surya Prakash · 1 year, 10 months ago

@Aditya Kumar – \(\text{Solution to Problem 4}\)

Clearly \(P(x)=x^2-x+1\) and \(P(1-x)=P(x)\)

Also \(\tan^{-1} (x)=\cot^{-1} \left(\frac{1}{x}\right)\)

\(\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}\)

\(\large{I=\displaystyle \int^{1}_{0} \left( \tan^{-1}(P(x)) \cot^{-1} \left( \sqrt{\frac{x}{1-x}}\right) \right) dx}\)

\(\large{2I=\frac{\pi}{2} \displaystyle \int^{1}_{0}\tan^{-1} (x^2-x+1) dx}\)

\(\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} - \displaystyle \int^{1}_{0} \tan^{-1}\left(\frac{1}{1+x(x-1)}\right)\right)}\)

\(\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} - \displaystyle \int^{1}_{0} \tan^{-1}\left(\frac{x-(x-1)}{1+x(x-1)}\right) dx \right)}\)

\(\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} -( \displaystyle \int^{1}_{0} \tan^{-1} (x) dx- \displaystyle \int^{1}_{0} \tan^{-1} (x-1) dx)\right)}\)

\(\large{2I=\frac{\pi}{2}\left(\frac{\pi}{2} -\left( \frac{\pi}{2}-2 \ln (2)\right)\right)}\)

\(\large{I=\frac{\pi}{2} \ln(2)}\) – Tanishq Varshney · 1 year, 10 months ago

@Surya Prakash – Another interesting elementary solution :

Lemma :

\[\large{\displaystyle \int_{0}^{\infty} \frac{\cos mx}{x^2+a^2} \mathrm{d} x=\frac{\pi}{2a} e^{-am}}\]

Proof :

\( \text{I}(m) = \displaystyle \int_{0}^{\infty} \dfrac{\cos mx}{x^2+a^2} \mathrm{d} x \tag{1} \)

Using Integration by parts,

\( \text{I}(m) = \displaystyle \int_{0}^{\infty} \dfrac{2x\sin mx}{m(x^2+a^2)^2} \mathrm{d} x \)

\(\implies m \cdot \text{I}(m) = \displaystyle \int_{0}^{\infty} \dfrac{2x \sin mx}{(x^2+a^2)^2} \mathrm{d} x \tag {2}\)

Partially Differentiating \((2)\) w.r.t. \(m\), we have,

\(\text{I} + m\dfrac{\partial}{\partial m} \text{I} = \displaystyle \int_{0}^{\infty} \dfrac {2x^2\cos(mx)}{(x^2+a^2)^2} \mathrm{d}x = \displaystyle \int_{0}^{\infty} \dfrac {2\cos(mx)}{(x^2+a^2)} \mathrm{d}x - 2a^2\displaystyle \int_{0}^{\infty}\dfrac{\cos mx}{(x^2+a^2)^2} \mathrm{d}x \)

\(\implies a\dfrac{\partial}{\partial m} \text{I} = \text{I} - 2b^2\displaystyle \int_{0}^{\infty}\dfrac{\cos mx}{(x^2+a^2)^2} \mathrm{d}x \tag{3}\)

Again, partially differentiating \((3)\) w.r.t. \(m\), we have,

\(m\dfrac{\partial^2}{\partial m^2} \text{I} = a^2 \displaystyle \int_{0}^{\infty} \dfrac{2x \sin mx}{(x^2+a^2)^2} \mathrm{d} x\)

\(\implies \dfrac{\partial^2}{\partial m^2} \text{I} = a^2 \text{I} \tag{*}\) (from \((2)\))

Note that \(I(0) = \dfrac{\pi}{2a}\) and \(I(\infty) = 0\)

Now, solving \((*)\), we have,

\(I = \dfrac{\pi}{2a} e^{-am}\) – Ishan Singh · 1 year, 9 months ago

@Surya Prakash – Solution to \(Problem ~15\)

Lemma

\(\large{\displaystyle \int^{\infty}_{0} \frac{\cos mx}{x^2+a^2} dx=\frac{\pi}{2a} e^{-am}}\)

\(\text{Proof}\)

Consider the integral \(\large{\displaystyle \int _{ C } f\left( z \right) dz}\) where \(\large{f(z)=\frac{e^{imz}}{z^2+a^2}}\) taken round the closed contour \(C\) consisting of the upper half of a large circle \(|z|=R\) and the real axis from \(-R\) to \(R\).

Poles of \(f(z)\) are given by

\(z^2+a^2=0 \rightarrow z=\pm ai\)

The only pole which lies within the contour is at \(z=ai\). The residue of \(f(z)\) at \(z=ai\)

\(\large{= \displaystyle \lim_{z \to ai} \frac{(z-ai)e^{imz}}{z^2+a^2}= \displaystyle \lim_{z \to ai} \frac{e^{imz}}{z+ai}=\frac{e^{-am}}{2ai}}\)

Hence by cauchy residue theorem

\(\large{\displaystyle \int _{ C } f\left( z \right) dz=2 \pi i \times \text{sum of residues}}\)

\(\large{\displaystyle \int _{ C } f\left( z \right) dz=2 \pi i \times \frac{e^{-am}}{2ai} }\)

\(\large{\displaystyle \int _{ C }\frac{e^{imz}}{z^2+a^2} dz \rightarrow \displaystyle \int^{R} _{ -R }\frac{e^{imx}}{x^2+a^2} dx= \frac{ \pi e^{-am}}{a}}\)

Equating Real parts

\(\large{\displaystyle \int^{\infty} _{ -\infty }\frac{\cos mx}{x^2+a^2} dx= \frac{ \pi e^{-am}}{a}}\)

\(\large{\Longrightarrow \displaystyle \int^{\infty} _{ 0 }\frac{\cos mx}{x^2+a^2} dx= \frac{ \pi e^{-am}}{2a}}\)

Now in the integral put \(\tan \theta =t\)

\(\large{\displaystyle \int^{\infty} _{ - \infty }\frac{\cos xt}{t^2+1} dt=\pi e^{-x}}\)

I would like to see the solution which doesn't involve contour!! – Tanishq Varshney · 1 year, 10 months ago

@Surya Prakash – Lemma: \[\int_{-\infty}^{\infty} \dfrac{\cos (xt)}{1+t^2} dt = \int_{-\infty}^{\infty} \dfrac{e^{-ixt}}{1+t^2} dt\]

Try to prove this yourselves.

Consider,

\[f(t) = \int_{-\infty}^{\infty} e^{-|x|} e^{ixt}dx = \dfrac{2}{1+t^2}\]

Above integral is easy to evaluate.

So,

\[f(t)=\dfrac{2}{1+t^2}\]

Now by Fourier inversion formula,

\[e^{-|x|} = \dfrac{1}{2\pi} \int_{-\infty}^{\infty} f(t) e^{-ixt} dt\]

So we get that

\[\int_{-\infty}^{\infty} \dfrac{\cos (xt)}{1+t^2} dt = \int_{-\infty}^{\infty} \dfrac{e^{-ixt}}{1+t^2} dt = \pi e^{-|x|}\] – Surya Prakash · 1 year, 10 months ago

@Tanishq Varshney – \[\int_{0}^{1} \dfrac{(1-x)}{\ln (x)} ( x + x^2 + x^{2^2} + x^{2^3} + \ldots ) = \sum_{n=0}^{\infty} \int_{0}^{1} \dfrac{x^{2^n} - x^{2^n + 1}}{\ln (x)} dx\]

Let \(f(a) = \int_{0}^{1} \dfrac{x^a}{\ln x} dx\). Differentiate it w.r.t \(a\). We get

\[f'(a) = \int_{0}^{1} x^a dx = \dfrac{1}{a+1}\]

So,

\[\int_{0}^{1} \dfrac{x^{2^n} - x^{2^n + 1}}{\ln(x)} dx = \int_{2^n+1}^{2^n} f'(a) da = \ln \left(\dfrac{2^n + 1}{2^n + 2} \right)\]

\[\int_{0}^{1} \dfrac{x^{2^n} - x^{2^n + 1}}{\ln(x)} dx = \ln (2^n +1) - \ln (2^n +2)\]

\[\sum_{n=0}^{\infty} \int_{0}^{1} \dfrac{x^{2^n} - x^{2^n + 1}}{\ln(x)} dx = \lim_{N \rightarrow \infty} \sum_{n=0}^{N} \ln (2^n +1) - \ln (2^n +2) = \lim_{N \rightarrow \infty} \sum_{n=0}^{N} \ln (2^n +1) -\ln (2) - \ln (2^{n-1} +1) \]

So, we have got a telescoping series which sums up to

\[= \lim_{N \rightarrow \infty} -(N+1)\ln(2) - \ln(3/2) + \ln (2^N + 1) = -\ln(3)\] – Surya Prakash · 1 year, 10 months ago

@Aditya Kumar – \(\large{Solution ~to~Problem ~12:}\)

\(\Large{f(x)=\frac{\sin (2 t \arctan \sqrt{x})}{\sqrt{x} (1+x)^t}=\displaystyle \sum^{\infty}_{k=0} \frac{\Gamma(2t+2k+1) \Gamma(k+1)}{\Gamma(2t) \Gamma(2k+2)}.\frac {(-x)^{k}}{k!}}\)

Using Ramanujan master theorem

\(\Large{\displaystyle \int^{\infty}_{0} x^{s-1} f(x)=\Gamma(s) \phi(-s)}\)

where, \(\phi(k)=\frac{\Gamma(2t+2k+1) \Gamma(k+1)}{\Gamma(2t) \Gamma(2k+2)}\)

\(x \rightarrow \tan^2 \theta\)

\(\Large{\displaystyle \int^{\frac{\pi}{2}}_{0} sin^{a-1} \theta (\cos \theta )^{2t-a} \cos (2t \theta) d \theta =\frac{\pi \Gamma(2t-a)}{2 \sin \left( \frac{\pi a}{2} \right) \Gamma(2t) \Gamma(-a)}}\) – Tanishq Varshney · 1 year, 10 months ago

@Pi Han Goh – Solution to Problem 11:

Pre-requisites:

\(I=\int \frac { -a\ln { \left( a+1 \right) } }{ \left( a-1 \right) \left( a^{ 2 }+1 \right) } da\\ { Apply\: Integral\: Substitution }:\quad \\ \int f\left( g\left( x \right) \right) \cdot g^{ ' }\left( x \right) dx=\int f\left( u \right) du,\: \quad u=g\left( x \right) \\ u=a\ln { \left( a+1 \right) } :\\ u=\frac { a }{ a+1 } +\ln { \left( a+1 \right) } da,\\ da=\frac { 1 }{ \frac { a }{ a+1 } +\ln { \left( a+1 \right) } } du\\ I=-\int \frac { u }{ \left( a-1 \right) \left( a^{ 2 }+1 \right) } \frac { 1 }{ \frac { a }{ a+1 } +\ln { \left( a+1 \right) } } du\\ I=-\frac { 1 }{ \left( a-1 \right) \left( a^{ 2 }+1 \right) \left( \frac { a }{ a+1 } +\ln { \left( a+1 \right) } \right) } \frac { u^{ 2 } }{ 2 } \\ On\quad substitution\quad \quad \\ I=-\frac { a^{ 2 }\ln { ^{ 2 } } \left( a+1 \right) }{ 2\left( a-1 \right) \left( a^{ 2 }+1 \right) \left( \frac { a }{ a+1 } +\ln { \left( a+1 \right) } \right) } \)

\(I\left( a \right) =\int _{ 0 }^{ 1 }{ \frac { ln\left( 1+ax \right) }{ \left( 1+x \right) \left( 1+{ x }^{ 2 } \right) } dx } \\ { I }^{ ' }\left( a \right) =\int _{ 0 }^{ 1 }{ \frac { x }{ \left( 1+x \right) \left( 1+ax \right) \left( 1+{ x }^{ 2 } \right) } dx } \\ On\quad splitting\quad as\quad partial\quad fractions\quad and\quad solving,\\ { I }^{ ' }\left( a \right) =\frac { -aln(a) }{ (a-1)({ a }^{ 2 }+1) } +\frac { ln2 }{ 2(a+1) } +\frac { ln2(1-a) }{ 4({ a }^{ 2 }+1) } \\ I\left( a \right) =\int _{ 0 }^{ 1 }{ \left( \frac { -aln(a) }{ (a-1)({ a }^{ 2 }+1) } +\frac { ln2 }{ 2(a+1) } +\frac { ln2(1-a) }{ 4({ a }^{ 2 }+1) } \right) da } \\ Now\quad on\quad solving\quad and\quad substituting\quad a=1,\quad we\quad get\\ I=\frac { 1 }{ 192 } \left( 36{ \left( ln2 \right) }^{ 2 }+12\pi \left( ln2 \right) -{ \pi }^{ 2 } \right) \) – Aditya Kumar · 1 year, 10 months ago

@Sudeep Salgia – Solution to Problem 6:

\(I=\int _{ a }^{ b }{ \text{arccos}\left( \frac { x }{ \sqrt { \left( a+b \right) x-ab } } \right) dx } \\ I=\int _{ a }^{ b }{ \text{arccos}\left( \frac { \frac { x }{ \sqrt { ab } } }{ \sqrt { \left( \frac { \left( a+b \right) }{ ab } \right) x-1 } } \right) dx } \\ \sqrt { \left( \frac { \left( a+b \right) }{ ab } \right) x-1 } =t\\ I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { 2t }{ a+b } arccos\left( \frac { \left( { t }^{ 2 }+1 \right) \sqrt { ab } }{ t } \right) \right) dt } \\ I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { 2t }{ a+b } arccos\left( \left( { t }+\frac { 1 }{ t } \right) \sqrt { ab } \right) \right) dt } \)

Now using Integration By Parts,

\(I=\int _{ \sqrt { \frac { a }{ b } } }^{ \sqrt { \frac { b }{ a } } }{ \left( \frac { { t }^{ 2 }+1 }{ { \left( \sqrt { \frac { b }{ a } } -\sqrt { \frac { a }{ b } } \right) }^{ 2 }-{ \left( { t }-\frac { 1 }{ t } \right) }^{ 2 } } \right) dt } \\ t=p+\sqrt { 1+{ p }^{ 2 } } \\ I=\int _{ \frac { 1 }{ 2 } \left( -\sqrt { \frac { b }{ a } } +\sqrt { \frac { a }{ b } } \right) }^{ \frac { 1 }{ 2 } \left( \sqrt { \frac { b }{ a } } -\sqrt { \frac { a }{ b } } \right) }{ \left( \frac { { 2p }^{ 2 } }{ \sqrt { \frac { \frac { a }{ b } +\frac { b }{ a } +2 }{ 4 } -{ p }^{ 2 } } } \right) dp } \)

Now,

\({ p }^{ 2 }=\left( \frac { \frac { a }{ b } +\frac { b }{ a } +2 }{ 4 } \right) sin\theta \)

This can be evaluated easily.

\(\therefore \quad I=\frac { \pi }{ 4 } \frac { { \left( b-a \right) }^{ 2 } }{ \left( a+b \right) } \)

Moral of the question:
To whatever extent we learn anything, we must never forget our basics.
Note that this question is solved using very basic concepts.

Any elegant modification to the solution will be accepted. – Aditya Kumar · 1 year, 10 months ago

@Tanishq Varshney – Solution to Problem 3: \(\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( tanx \right) }^{ \frac { 3 }{ 5 } } } dx=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( sinx \right) }^{ \frac { 3 }{ 5 } }{ \left( cosx \right) }^{ \frac { -3 }{ 5 } } } dx\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { B\left( \frac { 4 }{ 5 } ,\frac { 1 }{ 5 } \right) }{ 2 } ...........\left\{ B(x,y)\quad is\quad beta\quad function \right\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \Gamma \left( \frac { 4 }{ 5 } \right) \Gamma \left( \frac { 1 }{ 5 } \right) }{ 2\Gamma \left( 1 \right) } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { \pi }{ 2sin\left( \frac { \pi }{ 5 } \right) } \quad ...............\{ by\quad euler's\quad reflection\quad formula\} \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\frac { 2\pi }{ \sqrt { 10-2\sqrt { 5 } } } \) – Aditya Kumar · 1 year, 10 months ago

@Tanishq Varshney – Substitute \(t \to \sin(x)\). The integral becomes:

\[\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \ln { t } \ln { \left( 1-t^{ 2 } \right) } \left( \frac { 1 }{ t } \right) } dt\]

Substitute in the Taylor series for \(\ln (1-x)\):

\[-\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \ln { t } \sum _{ n=1 }^{ \infty } \frac { t^{ 2n-1 } }{ n } } dt=-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \int _{ 0 }^{ 1 }{ \ln { (t) } t^{ 2n-1 } } dt\]

Finally, integrate by parts to obtain

\[-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \frac { -1 }{ 4{ n }^{ 2 } } =\frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ { n }^{ 3 } } \] – Julian Poon · 1 year, 9 months ago

@Tanishq Varshney – Alternative approach could be (i know its too late :P) could be considering Beta function \((\dfrac{\Gamma(a) \Gamma(b)}{ \Gamma(a+b)})\) in trignometric form and differentiating it with respect to a and b and then putting a = 0 and b = 1. – Harsh Shrivastava · 1 year, 8 months ago

@Tanishq Varshney – Alternate Solution to problem 19 :

Lemma: \( \displaystyle \int _{ 0 }^{ \infty }{ \frac { { x }^{ m-1 }{ ln }^{ 2 }(x) }{ 1+{ x }^{ n } } dx } ={ \left(\frac { \pi }{ n } \right) }^{ 3 }\left(\csc { \left(\frac { m\pi }{ n } \right) } \cot ^{ 2 }{ \left(\frac { m\pi }{ n } \right) } +\csc ^{ 3 }{ \left(\frac { m\pi }{ n } \right) } \right)\quad \)

Proof : We begin with identity \(\displaystyle I = \int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ n } } } =\frac { \pi }{ n } \csc { (\frac { \pi }{ n } ) } \)

To prove this put \( \displaystyle y = \dfrac{1}{1+{x}^{n}} \) to get :

\(\displaystyle I =\dfrac{1}{n} \int _{ 0 }^{ 1 }{ { y }^{ -1/n }{ (1-y) }^{ 1/n-1 }dy } \)

Use definition of beta function and euler reflection formula to get :

\(\displaystyle I = \dfrac{1}{n}\int _{ 0 }^{ 1 }{ { y }^{ -1/n }{ (1-y) }^{ 1/n-1 }dy } =\dfrac{1}{n}\frac { \Gamma (1/n)\Gamma (1-1/n) }{ \Gamma (1) } =\frac { \pi }{ n } \csc { (\frac { \pi }{ n } ) } \)

Now consider the integral \( \displaystyle J(m) = \int _{ 0 }^{ \infty }{ \frac { { x }^{ m-1 }dx }{ 1+{ x }^{ n } } } \)

Put \({x}^{m}=y \) to get :

\( \displaystyle J(m) = \frac { 1 }{ m } \int _{ 0 }^{ \infty }{ \frac { dy }{ 1+{ y }^{ n/m } } } \)

Use the identity to get :

\( \displaystyle J(m) = \frac { \pi }{ n } \csc { (\frac { m\pi }{ n } ) } \)

Differentiate it two times with respect to \( m \) to prove the lemma :

My initial steps are the same as surya prakash and I directly get till :

\( \displaystyle I = \frac { \sqrt { 3 } }{ 2 } \int _{ 0 }^{ 1 }{ \frac { (1+y){ ln }^{ 2 }(y) }{ 1+{ y }^{ 3 } } dy } \)

Put \( \displaystyle y=\dfrac{1}{x} \) to get :

\( \displaystyle I = \frac { \sqrt { 3 } }{ 2 } \int _{ 1 }^{ \infty }{ \frac { (1+x){ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx } \)

Adding these two forms we have :

\( \displaystyle I = \frac { \sqrt { 3 } }{ 4 } \int _{ 0 }^{ \infty }{ \frac { (1+x){ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx } \)

\( \displaystyle I = \frac { \sqrt { 3 } }{ 4 } (\int _{ 0 }^{ \infty }{ \frac { { ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx } +\int _{ 0 }^{ \infty }{ \frac { x{ ln }^{ 2 }(x) }{ 1+{ x }^{ 3 } } dx } ) \)

Use the lemma to get :

\( \displaystyle I = \frac { \sqrt { 3 } }{ 4 } ({ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } )+{ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } ))=\frac { 5{ \pi }^{ 3 } }{ 81 } \) – Ronak Agarwal · 1 year, 9 months ago

@Tanishq Varshney – Lemma: \[\int_{0}^{1} x^{m} \ln ^{n} (x) dx = \dfrac{(-1)^n n!}{(m+1)^{n+1}}\]

Let \(y=\dfrac{\sin x}{\sin \left(x+\pi/3 \right)} \implies \tan x = \dfrac{\sqrt{3}y}{2-y} \implies dx = \dfrac{\sqrt{3}}{2} \dfrac{1}{y^2 - y +1}dy \).

So,

\[\begin{align*} \int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx &= \dfrac{\sqrt{3}}{2} \int_{0}^{1} \dfrac{\ln ^2 (y)}{y^2 - y +1} dy \\ &= \dfrac{\sqrt{3}}{2} \int_{0}^{1} \dfrac{(1+y)\ln ^2 (y)}{1+y^3} dy \end{align*}\]

Since, \(0 < y < 1\), we can use the expansion \(\dfrac{1}{1+y^3} = 1-y^3 + y^6 + \ldots = \sum_{k=0}^{\infty} (-1)^k y^{3k}\).

So, the integral becomes,

\[\begin{align*} \int_{0}^{1} \dfrac{(1+y)\ln ^2 (y)}{1+y^3} dy &= \int_{0}^{1} (1+y)\ln ^2 (y) \sum_{k=0}^{\infty} (-1)^k y^{3k} dy \\ &= \sum_{k=0}^{\infty} (-1)^k \Bigr[ \int_{0}^{1} y^{3k} \ln^2 (y) dy + \int_{0}^{1} y^{3k+1} \ln^2 (y) dy \Bigr] \\ &= 2 \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + 2 \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \end{align*}\]

So,

\[\int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx = \sqrt{3} \Bigr[ \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \Bigr] \]

Here comes the tricky part,

Consider the series \((0< q< 1)\)

\[\dfrac{1}{q^3} - \dfrac{1}{(1+q)^3} + \dfrac{1}{(2+q)^3} - \ldots\]

It is easy to prove that above series is equal to \(\dfrac{1}{8} \left( \zeta \left(3, \dfrac{q}{2} \right) - \zeta \left(3, \dfrac{q+1}{2} \right) \right)\), where \(\zeta(s,q)\) is Hurwitz Zeta Function.

Using the relation between Polygamma function and Hurwitz Zeta Function, We get

\[\dfrac{1}{q^3} - \dfrac{1}{(1+q)^3} + \dfrac{1}{(2+q)^3} - \ldots = \dfrac{1}{16} \left(\psi_{(2)} \left(\dfrac{q+1}{2} \right) - \psi_{(2)} \left(\dfrac{q}{2} \right) \right)\]

So, finally

\[\begin{align*} \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} &= \dfrac{1}{27} \left( \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1/3)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+2/3)^3} \right) \\ &= \dfrac{1}{27} \dfrac{1}{16} \left(\psi_{(2)} \left(\dfrac{1}{6}\right)-\psi_{(2)}\left(\dfrac{2}{3}\right)+\psi_{(2)}\left(\dfrac{1}{3}\right)-\psi_{(2)}\left(\dfrac{5}{6}\right) \right) = \dfrac{5 \sqrt{3} \pi^3}{243} \end{align*} \]

Above final calculation is done using Euler's Reflection Formula for polygamma function

So, finally

\[ \int_{0}^{\pi/3} \ln ^2 \left(\dfrac{\sin x}{\sin \left(\dfrac{\pi}{3} + x \right) } \right) dx = \sqrt{3} \Bigr[ \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+1)^3} + \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(3k+2)^3} \Bigr] = \sqrt{3} \dfrac{5 \sqrt{3}\pi^3}{243} = \dfrac{5 \pi^3}{81}\]

Sorry for missing the calculation part, As it became too lengthy to type here.

Notify me if there are any typing mistakes. – Surya Prakash · 1 year, 9 months ago

@Surya Prakash – Let \(I\) denote the value of the integral. And let \(3\tan (x) = 4\tan(y) \), differentiate with respect to \(x\): \[3\sec^2(x) = 4\sec^2(y) \left( \frac{dy}{dx} \right) \Rightarrow dx = \dfrac{4\sec^2(y)}{3\sec^2(x)} dy= \dfrac{12\sec^2(y)}{9\sec^2(x)} dy= \dfrac{12\sec^2(y)}{9\tan^2(x) + 9} dy= \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} dy \]

And of course, \(9\tan^2(x) + 16 = (3\tan(x))^2 + 16 = 16(\tan^2 (y) + 1) = 16\sec^2(y) \).

When \(x = 0\), \(y = 0\); when \(x\to \frac\pi2^- \), \(y \to \frac\pi2 ^-\) as well. The integral becomes

\[ \begin{eqnarray} I &=& \int_0^{\pi/2} \dfrac{1}{(16\sec^2(y))^3} \cdot \dfrac{12\sec^2(y)}{16\tan^2(y) + 9} \, dy \\ &=& \dfrac{12}{16^3} \int_0^{\pi/2} \dfrac1{\sec^4(y)} \cdot \dfrac1{16\tan^2 (y) + 9} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16\sin^2 (y) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16(1-\cos^2(y)) + 9\cos^2(y)} \, dy \\ &=& \dfrac{3}{1024} \int_0^{\pi/2} \dfrac{\cos^6(y)}{16 - 7\cos^2(y)} \, dy. \\ \end{eqnarray}\]

By long division, \( \dfrac{X^3}{16-7X} = -\dfrac17 X^2 - \dfrac{16}{49} X - \dfrac{4096}{343}\cdot \dfrac{1}{7X-16} - \dfrac{256}{343} \). We contine from above,

\[ \begin{eqnarray} \dfrac{1024}3 I &=& \int_0^{\pi/2} \bigg [ -\dfrac17 \cos^4(y) - \dfrac{16}{49} \cos^2 (y) - \dfrac{4096}{343} \cdot \dfrac1{7\cos^2(y) - 16} - \dfrac{256}{343} \bigg ] \, dy \\ \dfrac{1024}3 I &=& -\dfrac17\int_0^{\pi/2} \cos^4(y) \, dy - \dfrac{16}{49} \int_0^{\pi/2} \cos^2 (y) \, dy - \dfrac{8192}{343} \cdot \int_0^{\pi/2} \dfrac1{7(2\cos^2(y)) - 32} \, dy - \dfrac{128\pi}{343} \\ &=& -\left(\dfrac17 \cdot \dfrac{3!!}{4!!} \cdot \dfrac\pi2\right) - \left(\dfrac{16}{49}\cdot \dfrac{1!!}{2!!} \cdot \dfrac{\pi}2\right) - \dfrac{8192}{343} \int_0^{\pi/2} \dfrac1{7\cos(2y) - 25} \, dy -\left(\dfrac{128\pi}{343} \right) \\ &=& -\dfrac{2643\pi}{5488} + \dfrac{8192}{343} \int_0^{\pi/2}\dfrac1{25 - 7\cos(2y)} \, dy. \\ \end{eqnarray} \]

For the final integral, let \(t = \tan(y) \Rightarrow dy = \dfrac{dt}{t^2 + 1} \), and the integral becomes

\[\begin{eqnarray} \int_0^{\pi/2}\dfrac1{25 - 7\cos(2y)} \, dy &=& \int_0^\infty \dfrac{1}{25 - 7 \left( \frac{1-t^2}{1+t^2}\right) } \cdot \dfrac{dt}{t^2 + 1} \\ &=& \int_0^\infty \dfrac{t^2 + 1} {25 + 2t^2 - 7t + 7t^2} \cdot \dfrac{dt}{t^2 + 1} \\ &=& \int_0^\infty \dfrac{1}{32t^2 + 18} \, dt \\ &=& \dfrac1{32}\int_0^\infty \dfrac{1}{t^2 + (3/4)^2} \, dt \\ &=& \dfrac1{32} \cdot \left. \dfrac43 \tan^{-1} \left( \dfrac43 t \right) \right |_0^{t\to\infty} = \dfrac\pi{48}. \\ \end{eqnarray} \]

Simplify everything and we get the answer of \( I = \boxed{\dfrac{263 \pi}{5619712}} \approx 0.00014702512653568645897079 \).

As Surya Prakesh has pointed out, there's an easier method.

It can be easily proved that \(\int_{0}^{\pi /2} \dfrac{1}{\tan^2 (x ) + p} dx = \dfrac{\pi}{2(p+\sqrt{p})}\). Then differentiate it twice w.r.t \(p\). Then take \(p= \dfrac{4}{3}\). – Pi Han Goh · 1 year, 10 months ago

@Abhishek Bakshi – \(\large{\text{Solution to problem 8}}\)

\(t \rightarrow -t\)

adding this with the original equation

\(\Large{2f(w)=\frac{2}{\sqrt{2 \pi}} \displaystyle \int^{\infty}_{-\infty}e^{-a^2 t^2} \cos wt dt}\)

as \(e^{iwt}+e^{-iwt}=2 \cos wt=2 \Re(e^{iwt})\)

\(\Large{f(w)=\frac{1}{\sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-a^2 t^2}e^{iwt} dt \right)}\)

Let \(at=z\)

\(\Large{f(w)=\frac{1}{a \sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-z^2}e^{i\frac{w}{a}z} dz \right)}\)

\(\Large{f(w)=\frac{1}{a \sqrt{2 \pi}} \Re \left(\displaystyle \int^{\infty}_{-\infty}e^{-(z-\frac{iw}{2a})^2}e^{-\frac{w^2}{4a^2}} dz \right)}\)

Using Gaussian integral

\(\Large{f(w)=\frac{1}{a \sqrt{2}} e^{-\frac{w^2}{4a^2}}}\) – Tanishq Varshney · 1 year, 10 months ago

@Abhishek Bakshi – \[f(\omega) = \dfrac{1}{\sqrt{2 \pi}} \int_{- \infty} ^{\infty} e^{-a^2 t^2} e^{i\omega t}dt\]

It implies that \(f(\omega) = \big[\mathcal{F}(F) \big](\omega)\), where \(F(t) = e^{-a^2 t^2}\).

Now, differentiate \(F(t)\), we get \(F'(t) = -2a^2 t F(t)\).

Applying fourier transformation on both sides gives \(i \omega f(\omega) = -2i a^2 f'(\omega)\).

\[\implies \omega f(\omega) = -2a^2 \dfrac{d f(\omega)}{d \omega} \\ \implies f(\omega) = c e^{-\dfrac{\omega^{2}}{4a^{2}}}\]

Since \(c = f(0) =\dfrac{1}{\sqrt{2 \pi}} \int_{- \infty} ^{\infty} e^{-a^2 t^2} dt = 2 \dfrac{1}{\sqrt{2 \pi}} \int_{0} ^{\infty} e^{-a^2 t^2} dt =\dfrac{1}{a \sqrt{2 \pi}} \Gamma(1/2) = \dfrac{1}{a\sqrt{2}}\).

Therefore, \(f(\omega) =\dfrac{1}{a\sqrt{2}} e^{-\dfrac{\omega^{2}}{4a^{2}}}\). – Surya Prakash · 1 year, 10 months ago

@Aditya Kumar – Solution to Problem 7:

Lemma:

\(\Gamma \left( x \right) =\frac { { e }^{ -\gamma x } }{ x } \prod _{ n=1 }^{ \infty }{ { \left( 1+\frac { x }{ n } \right) }^{ -1 }{ e }^{ \frac { x }{ n } } } \)

Proof:

\(\Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { n }^{ x } }{ x } \prod _{ i=1 }^{ n }{ \left( \frac { i }{ x+i } \right) } \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { n }^{ x } }{ x } \prod _{ i=1 }^{ n }{ { \left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ xlog\left( n \right) } }{ x } \prod _{ i=1 }^{ n }{ { \left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ \sum _{ i=1 }^{ \infty }{ \left( \frac { x }{ n } \right) -\sum _{ i=1 }^{ \infty }{ \left( \frac { x }{ n } \right) } + } xlog\left( n \right) } }{ x } \prod _{ i=1 }^{ n }{ { \left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \\ But,\quad \gamma =\lim _{ n\rightarrow \infty }{ \left( \sum _{ i=1 }^{ \infty }{ \left( \frac { 1 }{ n } \right) } -log\left( n \right) \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ \sum _{ i=1 }^{ \infty }{ \left( \frac { x }{ n } \right) -\gamma x } } }{ x } \prod _{ i=1 }^{ n }{ { \left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \\ \Gamma \left( x \right) =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ -\gamma x } }{ x } \prod _{ i=1 }^{ n }{ { { e }^{ \left( \frac { x }{ n } \right) }\left( 1+\frac { x }{ i } \right) }^{ -1 } } \right) } \)

Now taking log on both sides and replacing x by x+1, we get

\(\log { \left( \Gamma \left( 1+x \right) \right) } =-\gamma x+\sum _{ k=2 }^{ \infty }{ \left( \frac { \zeta \left( k \right) }{ k } { \left( -x \right) }^{ k } \right) } \)

Now plugging these values into the question and by using Ramanujan's master theorem, we get

\[\int _{ 0 }^{ \infty }{ \left[ \frac { { x }^{ v-1 }\left( \gamma x+\log \Gamma \left( 1+x \right) \right) }{ { x }^{ 2 } } \right] \, dx } =\frac { \pi }{ \sin \left( \pi \left( v \right) \right) } \cdot \frac { \zeta \left( 2-v \right) }{ 2-v } \] – Aditya Kumar · 1 year, 10 months ago

@Aditya Kumar – Alternate Solution to Problem 7

Lemma : \( \displaystyle \int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx } =\Gamma (n+1)\zeta (n+1) \)

Proof :\( \displaystyle I = \int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx } =\int _{ 0 }^{ \infty }{ \frac { { x }^{ n }{ e }^{ -x } }{ 1-{ e }^{ -x } } dx } =\int _{ 0 }^{ \infty }{ { x }^{ n }\sum _{ r=1 }^{ \infty }{ { e }^{ -rx } } dx } =\sum _{ r=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -rx }dx } } \)

Note that I have used the series expansion of \( \dfrac{y}{1-y} \) and interchanged summation and integral.

\( \displaystyle I = \sum _{ r=1 }^{ \infty }{ \int _{ 0 }^{ \infty }{ { x }^{ n }{ e }^{ -rx }dx } } =\quad \sum _{ r=1 }^{ \infty }{ \frac { \Gamma (n+1) }{ { r }^{ n+1 } } } =\Gamma (n+1)\zeta (n+1) \)

I have here used the definitions of gamma function and the zeta function.

Now back to our problem.

Integrate it by parts to get :

\( \displaystyle I = \frac { { x }^{ v-2 } }{ v-2 } (\gamma x+\log {\Gamma (1+x) } ){ | }_{ 0 }^{ \infty }+\frac { 1 }{ 2-v } \int _{ 0 }^{ \infty }{ { x }^{ v-2 }(\gamma +\psi (1+x))dx } = \frac { 1 }{ 2-v } \int _{ 0 }^{ \infty }{ { x }^{ v-2 }(\gamma +\psi (1+x))dx } \)

Using the definition that \( \displaystyle \gamma +\psi (1+x)=\int _{ 0 }^{ 1 }{ \frac { 1-{ y }^{ x } }{ 1-y } dy } \) , we get :

\( \displaystyle I = \frac { 1 }{ 2-v } \int _{ 0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ { x }^{ v-2 }(\frac { 1-{ y }^{ x } }{ 1-y } )dxdy } } \)

Changing the order of integration we have :

\( \displaystyle I = \frac { 1 }{ 2-v } \int _{ 0 }^{ 1 }{ \frac { 1 }{ 1-y } \int _{ 0 }^{ \infty }{ { x }^{ v-2 }(1-{ y }^{ x })dxdy } } \)

Integrating it by parts we have :

\( \displaystyle I = \frac { 1 }{ 2-v } (\int _{ 0 }^{ 1 }{ \frac { dy }{ 1-y } (\frac { { x }^{ v-1 }(1-{ y }^{ x }) }{ v-1 } { | }_{ 0 }^{ \infty }+\frac { \ln(y) }{ v-1 } \int _{ 0 }^{ \infty }{ { x }^{ v-1 }{ y }^{ x }dx } ) } ) \)

\(\displaystyle I = \frac { 1 }{ (v-1)(2-v) } (\int _{ 0 }^{ 1 }{ \frac { \ln(y) dy }{ 1-y } (\int _{ 0 }^{ \infty }{ { x }^{ v-1 }{ y }^{ x }dx } ) } ) \)

\( \displaystyle I = \frac { 1 }{ (v-1)(2-v) } (\int _{ 0 }^{ 1 }{ \frac { \ln(y) dy }{ (1-y) } (\int _{ 0 }^{ \infty }{ { x }^{ v-1 }{ e }^{ -x(-\ln { (y)) } }dx } ) } ) \)

\( \displaystyle \Rightarrow I = \frac { 1 }{ (v-1)(2-v) } (\int _{ 0 }^{ 1 }{ \frac { \ln { (y) } dy }{ (1-y){ (-\ln { (y)) } }^{ v } } (\int _{ 0 }^{ \infty }{ { x }^{ v-1 }{ e }^{ -x }dx } ) } ) \)

\( \displaystyle \Rightarrow I = \frac { \Gamma (v) }{ (v-1)(2-v) } (\int _{ 0 }^{ 1 }{ \frac { \ln { (y) } dy }{ (1-y){ (-\ln { (y)) } }^{ v } } } ) \)

Put \(y={e}^{-x} \)

\( \displaystyle \Rightarrow I = \frac { \Gamma (v) }{ (1-v)(2-v) } (\int _{ 0 }^{ \infty }{ \frac { { x }^{ 1-v }dx }{ { e }^{ x }-1 } } ) \)

Using Lemma we have :

\( \displaystyle I = \frac { \Gamma (v)\Gamma (2-v)\zeta (2-v) }{ (1-v)(2-v) } =\frac { \Gamma (v)\Gamma (1-v)\zeta (2-v) }{ 2-v } \)

Using gamma reflection formula it becomes :

\( \displaystyle I = \frac { \pi \zeta (2-v) }{ \sin { (\pi v) } (2-v) } \)

Hence Proved. – Ronak Agarwal · 1 year, 9 months ago

@Surya Prakash – \[I=\int _{ 0 }^{ \pi }{ \frac { xsinx }{ { \left( { cos }^{ 2 }x+3 \right) }^{ 2 } } dx } \\ I=\frac { \pi }{ 2 } \int _{ 0 }^{ \pi }{ \frac { sinx }{ { \left( { cos }^{ 2 }x+3 \right) }^{ 2 } } dx } \\ cosx=t\quad \\ I=\frac { \pi }{ 2 } \int _{ -1 }^{ 1 }{ \frac { 1 }{ { \left( { t }^{ 2 }+3 \right) }^{ 2 } } dt } \\ t=\sqrt { 3 } tanx\\ I=\frac { \pi \sqrt { 3 } }{ 36 } \int _{ \frac { -\pi }{ 6 } }^{ \frac { \pi }{ 6 } }{ \left( cos2x+1 \right) dx } \\ \therefore I=\frac { \pi \sqrt { 3 } }{ 18 } \left\{ \frac { \sqrt { 3 } }{ 4 } +\frac { \pi }{ 6 } \right\} \]

In the first step, I used: \(\int _{ a }^{ b }{ f(x) } =\int _{ a }^{ b }{ f(a+b-x) } \) – Aditya Kumar · 1 year, 9 months ago

@Gautam Sharma – \[f(b) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1+ b \sin (x))}{\sin(x)} dx \\ f'(b) = \int_{-\pi/2}^{\pi/2}\dfrac{1}{1+ b\sin(x)} dx\]

Using Weierstrass Substitution i.e. \(t = \tan (x/2) \implies \sin(x) = \dfrac{2t}{1+t^2}\).

\[\begin{align*} f'(b) &= 2\int_{-1}^{1} \dfrac{dt}{t^2 + 2bt + 1}\\ &= 2\int_{-1}^{1} \dfrac{dt}{(t+b)^2 + (\sqrt{1-b^2})^2}\\ &= \dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{t+b}{\sqrt{1-b^2}} \right) \Bigr|_{-1}^{1}\\ &=\dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{1+b}{\sqrt{1-b^2}} \right) - \arctan\left( \dfrac{-1+b}{\sqrt{1-b^2}} \right)\Bigr] \\ &=\dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left( \dfrac{1+b}{\sqrt{1-b^2}} \right) + \arctan\left( \dfrac{1-b}{\sqrt{1-b^2}} \right)\Bigr] \\&= \dfrac{2}{\sqrt{1-b^2}} \Bigr[\arctan\left(\sqrt{\dfrac{1+b}{1-b}} \right) + \arctan \left(\sqrt{\dfrac{1-b}{1+b}} \right)\Bigr] \\&= \dfrac{\pi}{\sqrt{1-b^2}} \end{align*} \]

I used the fact that \(\arctan(x) + \arctan(1/x) = \pi/2\).

So,

\[f'(b) = \dfrac{\pi}{\sqrt{1-b^2}} \\ f(b) = \int \dfrac{\pi}{\sqrt{1-b^2}} db \\ f(b) = \pi \arcsin(b) +c\]

But \[f(b) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1+ b \sin (x))}{\sin(x)} dx \implies f(0) = \int_{-\pi/2}^{\pi/2} \dfrac{\ln ( 1)}{\sin(x)} dx =0\]

So, \(c=0\). Therefore, \(f(b) = \pi \arcsin(b)\). – Surya Prakash · 1 year, 9 months ago

@Surya Prakash – \(\text{Solution to problem 18}\)

\(\large{\displaystyle \int^{1}_{0} \frac{\ln^{2} (t)}{1+t^2} dt+\displaystyle \int^{\infty}_{1} \frac{\ln^{2} (t)}{1+t^2} dt}\)

In the latter integral \(t\rightarrow \frac{1}{t}\)

The whole expression becomes

\(\large{2 \displaystyle \int^{1}_{0} \frac{\ln^{2} (t)}{1+t^2} dt}\)

Now

\(\large{\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a } } dx=\frac { 1 }{ a+1 } \\ \displaystyle \int _{ 0 }^{ 1 }{ \frac { { \partial }^{ s } }{ { \partial x }^{ s } } } { x }^{ a }\quad dx=\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a } } { \ln { ^{ s } } { x } }dx=\frac { { \left( -1 \right) }^{ s }s! }{ { \left( a+1 \right) }^{ s+1 } } }\)

we have

\(\large{\frac{1}{1+t^2}=\displaystyle \sum^{\infty}_{r=0} (-1)^{r} t^{2r}}\)

\(\large{\longrightarrow 2 \displaystyle \int^{1}_{0} \displaystyle \sum^{\infty}_{r=0} \ln^{2} (t) (-1)^{r} t^{2r} dt}\)

\(\large{\longrightarrow 2 \displaystyle \sum^{\infty}_{r=0} (-1)^{r} \displaystyle \int^{1}_{0} t^{2r} \ln^{2} (t) dt}\)

\(\large{4 \displaystyle \sum^{\infty}_{r=0} \frac{(-1)^{r}}{(2r+1)^{3}}}\)

Now for the answer

Use of Dirichlet Beta function

which gives \(\large{4 \beta(3)=\frac{\pi^{3}}{8}}\) – Tanishq Varshney · 1 year, 9 months ago

@Pi Han Goh – \[\begin{align*} \int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx &=\int_{-\infty}^{\infty} \dfrac{x^3 + x^2}{x^5+1}dx \\ &= \int_{-\infty}^{0} \dfrac{x^3 + x^2}{x^5+1}dx + \int_{0}^{\infty} \dfrac{x^3 + x^2}{x^5+1} dx \\ &= \int_{0}^{\infty} \dfrac{x^2 - x^3}{1-x^5}dx + \int_{0}^{\infty} \dfrac{x^3 + x^2}{x^5+1} dx \\ &= 2 \int_{0}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx \end{align*}\]

\[\begin{align*} \int_{0}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx &= \int_{0}^{1} \dfrac{x^2 - x^8}{1-x^{10}} dx + \int_{1}^{\infty} \dfrac{x^2 - x^8}{1-x^{10}} dx \\ &= \int_{0}^{1} \dfrac{1+ x^2 - x^6 - x^8}{1-x^{10}}dx \end{align*}\]

I just took the substitution \(t=\dfrac{1}{x}\) in the above steps.

So,

\[\begin{align*} \int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx &=2 \int_{0}^{1} \dfrac{1+ x^2 - x^6 - x^8}{1-x^{10}}dx \\ &= \dfrac{1}{5} \int_{0}^{1} \dfrac{t^{-9/10} + t^{-7/10} - t^{-3/10} - t^{-1/10}}{1-t} dt \\ &= \dfrac{1}{5} \left( H_{-1/10} + H_{-3/10} - H_{-7/10} - H_{-9/10} \right) \\&=\dfrac{1}{5} ((\psi(9/10) - \psi(1/10)) + (\psi(7/10) - \psi(3/10))) \end{align*}\]

By euler reflection formula for Digamma function i.e. \(\psi(1-x) - \psi(x) = \pi \cot(\pi x)\), we get

\[(\psi(9/10) - \psi(1/10)) + (\psi(7/10) - \psi(3/10)) = \pi \cot(\pi/10) + \pi \cot(3\pi/10)\]

So,

\[\int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx = \dfrac{1}{5} ((\psi(9/10) - \psi(1/10)) + (\psi(7/10) - \psi(3/10))) = \dfrac{\pi}{5}\cot(\pi/10) + \dfrac{\pi}{5} \cot(3\pi/10)\]

Similarly, evaluating the integral in the denominator, we get

\[ \int_{-\infty}^{\infty}\dfrac{x}{x^4 - x^3 + x^2 - x+1} dx = \dfrac{2\pi}{5} \cot (3\pi /10)\]

So finally

\[\left(\int_{-\infty}^{\infty}\dfrac{x^2}{x^4 - x^3 + x^2 - x+1} dx \right) \div \left(\int_{-\infty}^{\infty}\dfrac{x}{x^4 - x^3 + x^2 - x+1} dx \right) \\ = \dfrac{\dfrac{\pi}{5}\cot(\pi/10) + \dfrac{\pi}{5} \cot(3\pi/10)}{\dfrac{2\pi}{5} \cot (3\pi /10)} = \dfrac{\cot(\pi/10)+\cot(3\pi/10)}{2 \cot (3\pi /10)}= \dfrac{3+\sqrt{5}}{2}\]

Note: \(\cot (\pi/10) = \sqrt{5+2\sqrt{5}}\) and \(\cot (3 \pi/10) = \sqrt{5-2\sqrt{5}}\). – Surya Prakash · 1 year, 9 months ago

@Tanishq Varshney – This is very similar to Problem 5. Integrate by Parts: with \(dv = \dfrac x{1+x^4} \, dx, u = \cos^{-1}(x) \), then \(du =- \dfrac1{\sqrt{1-x^2}} dx, v = \dfrac12 \tan^{-1}(x^2) \).

\[\begin{eqnarray} \int u \, dv &= &uv - \int v \, du \\ &=& \require{cancel} \cancelto{0}{ \left . \cos^{-1}(x) \cdot \dfrac12 \tan^{-1} (x^2) \right |_{x=0}^{x=1}}+ \dfrac12 \int_0^1 \dfrac {\tan^{-1}(x^2)}{\sqrt{1-x^2}} \, dx \quad\quad && \text{ let } x = \sin(y) \Rightarrow dy = \dfrac{dx}{\sqrt{1-x^2}} \\ &=& \dfrac12 \int_0^{\pi /2} \tan^{-1} (\sin^2 (y)) \, dy \quad\quad && \text{ let } z = \dfrac \pi2 - y \\ &=& \dfrac12 \int_0^{\pi /2} \tan^{-1} (1 \cdot \cos^2 (z)) \, dz \quad\quad && \text{ refer to solution to problem 5} \\ &=& \dfrac12 \dfrac \pi2 \tan^{-1} \left ( \sqrt{\dfrac{\sqrt{1^2+1}+1}{2} }\right) \\ &=& \dfrac\pi4 \tan^{-1} \left( \sqrt{\dfrac{\sqrt2-1}{2}}\right) \approx 0.335426737379 \\ \end{eqnarray}\] – Pi Han Goh · 1 year, 10 months ago

@Surya Prakash – Solution to problem 20

Let \[(1+x)=t \\ I=\large \int \dfrac{dt}{t^3 +1}\]

By partial fractions

\[I=\int \frac { 2-t }{ 3({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ 2({ t }^{ 2 }-t+1) } dt-\frac { 2t-1 }{ 6({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ 2({ (t-\frac { 1 }{ 2 } ) }^{ 2 }+\frac { 3 }{ 4 } ) } dt-\frac { 2t-1 }{ 6({ t }^{ 2 }-t+1) } dt+\frac { 1 }{ 3(t+1) } dt \\ I=\int \frac { 1 }{ \sqrt { 3 } } \tan ^{ -1 }{ \frac { 2t-1 }{ \sqrt { 3 } } } -\frac { 1 }{ 6 } \ln({ t }^{ 2 }-t+1)+\frac { 1 }{ 3 } \ln(t+1)+c \\ I=\int \frac { 1 }{ \sqrt { 3 } } \tan ^{ -1 }{ \frac { 2x+1 }{ \sqrt { 3 } } } +\frac { 1 }{ 6 } \ln\left(\frac { { (x+2) }^{ 2 } }{ { x }^{ 2 }+x+1 } \right)+c\]

Putting limits and evaluating we get \(\frac{\pi}{3\sqrt3}-\frac{\ln2}{3}\) – Gautam Sharma · 1 year, 9 months ago