Brilliant Integration Contest - Season 2 (Part-1)

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 \(ln2-1\)

\(\Large \text{Solution to Problem 1:}\)

\[\int_{0}^{1} \dfrac{x \ln(x)}{\sqrt{1-x^2}} dx\] 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 \(y= \dfrac{1}{2}\).

\[\int_{0}^{\pi / 2} \sin(z) \ln (\sin (z)) dz = \dfrac{1}{4} \dfrac{\Gamma (1) \Gamma(1/2) }{\Gamma (3/2)} (\psi (1) - \psi(3/2)) = \ln (2) -1\]

\(\large{Solution~to~Problem~ 2}\)

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

\(\large{\boxed{\frac{\sqrt{3} \Gamma \left(\frac{2}{3}\right)}{3.2^{\frac{4}{3}}}}}\)

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.

\(\text{Solution to Problem 4}\)

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

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

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

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

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

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

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

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

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

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

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.

\(\text{Solution to Problem 24}\)

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

Refer to problem 18 for this step.

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

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

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

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

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

\(\large{=\frac{\sqrt{10-\sqrt{2}}}{2} \pi}\)

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

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

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, \[\int \dfrac{x^2 + 20}{(x \sin(x) + 5 \cos(x))^2} dx = \int \dfrac{ u^2 \dfrac{d\left(\dfrac{v}{u}\right)}{dx}}{u^2} dx= \dfrac{v}{u} = \dfrac{5 \sin(x) - x \cos(x)}{x \sin(x) + 5 \cos(x)}\]

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,

\[\begin{align*} \int_{0}^{1} x \ln \left(\dfrac{1+x}{1-x} \right) dx &= \lim_{a \rightarrow 0} \int_{0}^{1-a} x \ln \left(\dfrac{1+x}{1-x} \right) dx \\ &= \lim_{a \rightarrow 0} \dfrac{(1-a)^2}{2} \ln \left(\dfrac{2-a}{a} \right) + 1 - a - \dfrac{1}{2} \ln \left(2-a \right) + \dfrac{1}{2} \ln a \\ &= 1\end{align*}\]

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.

\(\large{\text{Solution to problem 8}}\)

\(t \rightarrow -t\)

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

\(\Large{f(w)=\frac{1}{a \sqrt{2}} e^{-\frac{w^2}{4a^2}}}\)

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

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, \(f(\omega) =\dfrac{1}{a\sqrt{2}} e^{-\dfrac{\omega^{2}}{4a^{2}}}\).

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 \(p= \dfrac{4}{3}\).

Solution to Problem 11:

Pre-requisites:

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

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

\(\large{Solution ~to~Problem ~12:}\)

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

Using Ramanujan master theorem

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

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

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

\(\Large{\displaystyle \int^{\frac{\pi}{2}}_{0} sin^{a-1} \theta (\cos \theta )^{2t-a} \cos (2t \theta) d \theta =\frac{\pi \Gamma(2t-a)}{2 \sin \left( \frac{\pi a}{2} \right) \Gamma(2t) \Gamma(-a)}}\)

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

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

\[= \lim_{N \rightarrow \infty} -(N+1)\ln(2) - \ln(3/2) + \ln (2^N + 1) = -\ln(3)\]

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,

\(I = \dfrac{\pi}{2a} e^{-am}\)

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!!

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

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

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

\[\int _{ 0 }^{ \infty }{ \left[ \frac { { x }^{ v-1 }\left( \gamma x+\log \Gamma \left( 1+x \right) \right) }{ { x }^{ 2 } } \right] \, dx } =\frac { \pi }{ \sin \left( \pi \left( v \right) \right) } \cdot \frac { \zeta \left( 2-v \right) }{ 2-v } \]

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.

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 :

\( \displaystyle I = \frac { \sqrt { 3 } }{ 4 } ({ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } )+{ (\frac { \pi }{ 3 } ) }^{ 3 }(\frac { 2 }{ \sqrt { 3 } } .\frac { 1 }{ 3 } +\frac { 8 }{ 3\sqrt { 3 } } ))=\frac { 5{ \pi }^{ 3 } }{ 81 } \)

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.

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

\[-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ n } \frac { -1 }{ 4{ n }^{ 2 } } =\frac { 1 }{ 8 } \sum _{ n=1 }^{ \infty } \frac { 1 }{ { n }^{ 3 } } \]

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.

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

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

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

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

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 \(\cot (3 \pi/10) = \sqrt{5-2\sqrt{5}}\).

\(\text{Solution to problem 18}\)

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

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 \(\large{4 \beta(3)=\frac{\pi^{3}}{8}}\)

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

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, \(f(b) = \pi \arcsin(b)\).

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

In the first step, I used: \(\int _{ a }^{ b }{ f(x) } =\int _{ a }^{ b }{ f(a+b-x) } \)

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 \(\frac{\pi}{3\sqrt3}-\frac{\ln2}{3}\)

I used complex analysis