Brilliant Integration Contest - Season 1

Here is the solution: \(\displaystyle I = \int \limits_0^a \frac{ \text{ln } x }{\sqrt{ax - x^2}} \text{d}x \)

\(\displaystyle I = \int \limits_0^a \frac{ \text{ln } (a-x) }{\sqrt{ax - x^2}} \text{d}x \)

\(\displaystyle \Rightarrow 2I = \int \limits_0^a \frac{ \text{ln } (ax - x^2) }{\sqrt{ax - x^2}} \text{d}x \)

Put \(\displaystyle x = \frac{a}{2} (1 - \sin t ) \Rightarrow ax - x^2 = \frac{a^2}{4} \cos ^2 t \)

Therefore, the integral becomes, \(\displaystyle 2I = \int \limits_{\frac{\pi}{2}}^{\frac{-\pi}{2}} \frac{ \text{ln } (\frac{a^2}{4} \cos ^2 t ) }{\sqrt{\frac{a^2}{4} \cos ^2 t }} (\frac{a}{2} \cos t )(-\text{d}t) \)

Rearranging, we obtain,
\(\displaystyle 2I = 4\int \limits_0^{\frac{\pi}{2}} \text{ln } (\frac{a}{2} \cos t ) \text{d}t \)

The value of \(\displaystyle \int \limits_0^{\frac{\pi}{2}} \text{ln } ( \cos \theta ) \text{d}\theta \) is \(\displaystyle \frac{ -\pi \text{ ln } 2 }{2} \) (which I have calculated separately and I can post if it is required).

Thus, we obtain, \(\displaystyle I = \pi \text{ ln } \bigg( \frac{a}{2} \bigg) - 2\frac{ \pi \text{ ln } 2 }{2} = \pi \text{ ln } \bigg(\frac{a}{4} \bigg)\)

Solution of Problem 3 :

Use integration by parts by taking \(u=\sin^2x\) and \(dv=\dfrac{dx}{x^2}\), we get

\[\begin{align} \int_0^\infty\frac{\sin^2x}{x^2}\,dx&=-\left.\frac{\sin^2x}{x}\right|_0^\infty+\int_0^\infty\frac{2\sin x\cos x}{x}\,dx\\ &=0+\int_0^\infty\frac{\sin 2x}{x}\,dx\\ &=\int_0^\infty\frac{\sin t}{t}\,dt\qquad\Rightarrow\qquad t=2x\\ \end{align}\]

Now consider

\[\begin{equation} I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt\qquad a\ge0 \end{equation}\]

so that \(I(\infty)=0\) and our considered integral is \(I(0)\). Differentiating w.r.t. \(a\) and then integrating back, we get

\[\begin{align} I'(a)&=-\int_0^\infty e^{-at}\sin t\,dt\\ &=-\frac{1}{1+a^2}\qquad\Rightarrow\qquad\text{integration by parts twice}\\ I(a)&=-\int \frac{1}{1+a^2}\,da\\ &=-\arctan(a)+C \end{align}\]

For \(I(\infty)=0\), implying \(C=\dfrac{\pi}{2}\). Hence

\[\begin{equation} I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt=\frac{\pi}{2}-\arctan(a) \end{equation}\]

and

\[\begin{equation} I(0)=\int_0^\infty\frac{\sin t}{t}\,dt=\frac{\pi}{2} \end{equation}\]

Thus

\[\begin{equation} \int_0^\infty\frac{\sin^2x}{x^2}\,dx=\frac{\pi}{2} \end{equation}\]

Problem 4 :

Prove

\[\begin{equation} \int_0^{\pi/2}\frac{\ln(\cos x)}{\sin x}\,dx=-\frac{\pi^2}{8} \end{equation}\]

Solution of Problem 7:

Proposition :

\[\begin{align} \int_0^{\pi} \frac{dx}{p+\cos x}\,dx=\frac{\pi}{\sqrt{p^2-1}} \end{align}\]

Proof :

It can be proven by using Weierstrass substitution: \(t=\tan\left(\dfrac{x}{2}\right)\), then

\[\begin{align} \\ \int_0^{\pi} \frac{dx}{p+\cos x}\,dx &=\int_0^{\infty} \frac{2}{p+1+(p-1)t^2}\,dt\qquad\Rightarrow\qquad t=\sqrt{\frac{p+1}{p-1}}\tan y\\ &=\left.\frac{2}{\sqrt{p^2-1}}\arctan t\;\right|_0^{\infty}\\ &=\frac{\pi}{\sqrt{p^2-1}}\qquad\qquad\text{Q.E.D.} \end{align}\]

Differentiating the proposition w.r.t. \(p\) twice and setting \(p=\sqrt{10}\), we have

\[\begin{align} \\ \frac{\partial^2}{\partial p^2}\int_0^\pi\frac{1}{p+\cos x}\, dx&=\frac{\partial^2}{\partial p^2}\left[\frac{\pi}{\sqrt{p^2-1}}\right]\\ \int_0^\pi\frac{2}{\left(p+\cos x\right)^3}\, dx&=\frac{\pi\left(2p^2+1\right)}{\sqrt{\left(p^2-1\right)^5}}\\ \int_0^\pi\frac{1}{\left(\sqrt{10}+\cos x\right)^3}\, dx&=\frac{7\pi}{162} \end{align}\]

Problem 8 :

Prove

\[\begin{equation} \int_0^{\Large\frac{\pi}{2}}\frac{dx}{1+8\sin^2(\tan x)}=\frac{\pi}{6}\left(\frac{2e^2+1}{2e^2-1}\right) \end{equation}\]

Solution to problem 10

First integrate by parts -

\[I= \int_{0}^{\infty} \dfrac{\sin^3{x}}{x^2} dx =\int_{0}^{\infty} \dfrac{3\sin^2{x} \cos{x}}{x} dx\]

Then, use the property of the laplace transform that \(\mathcal{L} \left\{\dfrac{f(t)}{t}\right\}(s)=\displaystyle\int_{s}^{\infty} F(p) dp\) (in our case, \(s \rightarrow 0\)).

\[I=3\int_{0}^{\infty} \mathcal{L} \{\sin^2{t} \cos{t}\}(p) \ dp\]

\[I=3\int_{0}^{\infty} \dfrac{2s}{(s^2+1)(s^2+9)} ds=\boxed{\dfrac{3\log {3}}{4}}\] The last integral can be found by using partial fractions.

Note : \(\mathcal{L}\{f(t)\}(s)=\displaystyle\int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t\)

Problem 11

Find

\[\int_{0}^{\infty} \dfrac{\log {x}}{x^2+\alpha^2} \mathrm{d}x \ \ \ \ \ \text{for} \ \alpha>0 \]

Solution of Problem 13 :

First we prove the following proposition:

Proposition :

\[\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}} \qquad\text{for }\ k=0,1,2,\ldots\]

Proof :

Note that \[ \int_0^1 x^a\ dx=\frac1{a+1}\qquad\text{for }\ \alpha>-1. \] Differentiating equation above \(k\) times w.r.t. \(a\) we have \[ \int_0^1 \frac{\partial^k}{\partial a^k}\left(x^a\right)\ dx=\int_0^1 x^a \ln^k x\ dx=\frac{(-1)^k\, k!}{(a+1)^{k+1}}\qquad\text{Q.E.D.} \]

Now, we will evaluate the general case of

\[\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\int_0^\infty\frac{x^{a-1}\,e^{-bx}}{1-e^{-bx}}\,dx\qquad\text{for }\ a,b>0\]

Set \(y=e^{-bx}\), then

\[\int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx=\frac{(-1)^{a-1}}{b^a}\int_0^1\frac{\ln^{a-1}y}{1-y}\,dy\]

Use a geometric series for \(\dfrac{1}{1-y}\) then interchange the integral and summation sign which is justified by the Fubini–Tonelli theorem and apply the previous proposition, we have

\[\begin{align} \int_0^\infty\frac{x^{a-1}}{e^{bx}-1}\,dx&=\frac{(-1)^{a-1}}{b^a}\int_0^1\sum_{k=0}^\infty y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty\int_0^1 y^k\,\ln^{a-1}y\,dy\\ &=\frac{(-1)^{a-1}}{b^a}\sum_{k=0}^\infty \frac{(-1)^{a-1}\, (a-1)!}{(k+1)^{a}}\\ &=\frac{(a-1)!}{b^a}\sum_{k=1}^\infty \frac{1}{k^{a}}\\ &=\frac{\Gamma(a)\,\zeta(a)}{b^a} \end{align}\]

where \(\Gamma(a)\) is the gamma function and \(\zeta(a)\) is the Riemann zeta function.

Hence, by setting \(a=n+1\) and \(b=1\), we have

\[\int_0^\infty\frac{x^{n}}{e^{x}-1}\,dx=\Gamma(n+1)\,\zeta(n+1)\]

Problem 14 :

Prove that

\[\begin{align} \int_{0}^1 \ln\left(\frac{3+x}{3-x}\right)\,\frac{dx}{\sqrt{x(1-x)}}=\pi\ln\left(\dfrac{7+4\sqrt{3}}{5+2\sqrt{6}}\right)\\ \end{align}\]

Since @Pranav Arora is unable to propose a problem (I hope it's only temporary) and to make this contest sustains, then according to rule 4, I, as the last solver, have a right to propose a new one. Here is the problem:

PROBLEM 9

Prove that

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

\(Solution\quad of\quad problem\quad 9\)

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

Put \(x=cos(\theta)\) to get our integral as :

\(\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln({ cot }^{ 2 }\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } } \)

\(\Rightarrow \displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(tan\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } } \)

Put \(tan(\theta/2)=x\) to get our integral as :

\(\displaystyle (-4)\int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } \)

I proved in my solution to problem 4 that :

\(\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } =\frac { -{ \pi }^{ 2 } }{ 8 } \)

Using this I get :

\(I=\frac{{\pi}^{2}}{2}\)

\(Problem\quad 10\)

Find \(\displaystyle \int _{ 0 }^{ \infty }{ \frac { { sin }^{ 3 }x }{ { x }^{ 2 } } dx } \)

Solution of Problem 11 :

Using substitution \(u=\dfrac{\alpha^2}{x}\;\Rightarrow\;x=\dfrac{\alpha^2}{u}\;\Rightarrow\;dx=-\dfrac{\alpha^2}{u^2}\ du\), then

\[\begin{align} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\int_0^\infty\frac{\ln \left(\dfrac{\alpha^2}{u}\right)}{\left(\dfrac{\alpha^2}{u}\right)^2+\alpha^2}\cdot \dfrac{\alpha^2}{u^2}\ du\\ I(\alpha)&=\int_0^\infty\frac{2\ln \alpha-\ln u}{\alpha^2+u^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-\int_0^\infty\frac{\ln u}{u^2+\alpha^2}\ du\\ &=2\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du-I(\alpha)\\ I(\alpha)&=\ln \alpha\int_0^\infty\frac{1}{\alpha^2+u^2}\ du. \end{align}\]

The last integral can easily be evaluated by using substitution \(u=\alpha\tan\theta\), then

\[\begin{align} \int_0^\infty\frac{\ln x}{x^2+\alpha^2}\ dx&=\frac{\ln \alpha}{\alpha}\int_0^{\Large\frac\pi2} \ d\theta\\ &=\frac{\pi\ln \alpha}{2\alpha} \end{align}\]

Problem 12 :

Prove

\[\begin{equation} \int_{-\infty}^{\infty}\frac{\cos \left(\, \arctan 2x\right)}{(1+x^2)\sqrt{1+4x^2}}\,dx=\frac{\pi}{3} \end{equation}\]

\(Solution\quad of\quad Problem\quad 12\)

It's a very good disguised integral.

First note that the function is an even function hence our integral can be written as :

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

Now we know that \(\displaystyle cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}\). Using this our integral becomes :

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

Splliting it by partial fractions we get :

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

\(\displaystyle I=\frac { 2 }{ 3 } (2{ tan }^{ -1 }(2x)\overset { \infty }{ \underset { 0 }{ | } } -{ tan }^{ -1 }(x)\overset { \infty }{ \underset { 0 }{ | } } )\)

\(\displaystyle I=\frac { \pi }{ 3 }\)

\(Problem\quad 13\)

Find closed form of \(\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx } \)

\(Solution\quad of\quad problem\quad 14\)

Lemma

\(\displaystyle\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx } =2\pi ln(\frac { a+b }{ 2 } )\)

Proof :

\(\displaystyle I(a)=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln({ a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x)dx }\)

Differentiating with respect to \(a\) we get :

\(\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 2 }x }{ { a }^{ 2 }{ cos }^{ 2 }x+{ b }^{ 2 }{ sin }^{ 2 }x } dx } \)

Put \(tan(x)=t\) to get our integral as :

\(\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } } \)

\(\displaystyle =2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } } =\frac { 2a }{ { b }^{ 2 }-{ a }^{ 2 } } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { { b }^{ 2 }dt }{ { a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 } } } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ 1+{ t }^{ 2 } } } )\)

Solving it we get :

\(\displaystyle {I}^{'}(a)=\frac { 2\pi }{ a+b } \)

\(\Rightarrow \displaystyle I(a)=2\pi ln(a+b)+C\)

Put \(a=b=1\) to get \(C=-2\pi ln(2)\)

Hence \(\displaystyle I(a)=2\pi ln(\frac{a+b}{2})\)

In our integral in the given question put \(x=\frac{1-sin\theta }{2}\) to get the integral as :

\(I= \displaystyle \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7-sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5+sin(\theta))dx } \)

Using the property \(\displaystyle \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(a+b-x)dx } \) we get :

\(\displaystyle I=\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(7+sin(\theta))dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(5-sin(\theta))dx } \)

Adding these two we get :

\(\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49-{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25-{ sin }^{ 2 }(\theta))dx } )\)

\(\displaystyle I=\frac { 1 }{ 2 } (\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(49{ cos }^{ 2 }\theta+48{ sin }^{ 2 }\theta)dx } -\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ ln(25{ cos }^{ 2 }\theta+24{ sin }^{ 2 }\theta)dx } )\)

Using our given lemma we get :

\(\displaystyle I=\frac { 1 }{ 2 } (2\pi ln(\frac{7+4\sqrt { 3 }}{2} )-2\pi ln(\frac{5+2\sqrt { 6 }}{2} ))=\pi ln(\frac { 7+4\sqrt { 3 } }{ 5+2\sqrt { 6 } } )\)

\(Problem\quad 15\)

Evaluate \(\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ ln(x) } +\frac { 1 }{ 1-x } dx } \)

_ Solution of problem 14 _

\(I = \displaystyle \int_{0}^{1} \ln \bigg(\dfrac{3+x}{3-x} \bigg) \dfrac{1}{\sqrt{x(1-x^2)}} {\mathrm dx} \)

= \(\displaystyle \int_{0}^{1} \ln(1 + x/3) \dfrac{1}{\sqrt{x(1-x)}} {\mathrm dx} \)

= \(\displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 + \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta - \displaystyle \int_{0}^{\pi/2} 2 \ln\bigg(1 - \frac{1}{3} \sin^2 \theta\bigg) {\mathrm d} \theta\)

Lemma \(\displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 \theta) = \pi \ln \bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)\)

Proof

Let \(I(a) = \displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 x) {\mathrm dx}\)

\(I'(a) = \displaystyle \int_{0}^{\pi/2} \dfrac{\sin^2 x}{1+ a \sin^2 x} {\mathrm dx}\)

\(I'(a) = \displaystyle \int_{0}^{\pi/2} \dfrac{1}{a}\bigg(1 - \dfrac{\sec^2 x}{1+(a+1) \tan^2x} {\mathrm dx}\bigg)\)

= \( \dfrac{\pi}{2a} - \dfrac{1}{a} \bigg(\displaystyle \int_{0}^{\infty} \dfrac{dz}{1+(a+1)z^2}\bigg)\)

=\(\dfrac{\pi}{2a} \bigg(1 - \dfrac{1}{\sqrt{a+1}}\bigg)\)

Hence, \(I(a) - I(0) = \displaystyle \int_{0}^{a} \dfrac{\pi}{2} \bigg(\dfrac{1}{x} - \dfrac{1}{x\sqrt{x+1}}\bigg) {\mathrm dx}\)

= \(\pi \ln(\sqrt{x+1} +1)\bigg|_{0}^{a} = \pi\ln\bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)\) (Note that \(I(0) = 0\))

Put \(a=1/3\) , and \(-1/3\), then substitute back in original integral to get:

\(I = 2 \pi \ln \bigg(\dfrac{1+\sqrt{4/3}}{1+\sqrt{2/3}} \bigg) = \boxed{\pi \ln \bigg(\dfrac{7 + 4\sqrt{3}}{5+2\sqrt{6}}\bigg)}\)

Solution of Problem 15 :

We have \[\begin{equation} \int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx \end{equation}\]

Proposition :

\[\begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s \end{equation}\]

Proof :

Let

\[\begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx \end{equation}\]

then

\[\begin{align} I'(s)&=\int_0^1 \frac{1-x^s}{1-x}\,dx\\ I''(s)&=-\int_0^1 \frac{x^s\ln x}{1-x} \,dx\\ &=-\int_0^1\sum_{n=0}^\infty x^{n+s}\ln x\,dx\\ &=-\sum_{n=0}^\infty\partial_s\int_0^1 x^{n+s}\,dx\\ &=-\sum_{n=0}^\infty\partial_s\left[\frac{1}{n+s+1}\right]\\ &=\sum_{n=0}^\infty\frac{1}{(n+s+1)^2}\\ &=\psi_1(s+1)\\ I'(s)&=\int\psi_1(s+1)\,ds\\ I'(s)&=\int\frac{\partial}{\partial s}\bigg[\psi(s+1)\bigg]\,ds\\ I'(s)&=\psi(s+1)+C\\ \end{align}\] For \(s=0\), we have \(I'(0)=0\). Implying \(C=-\psi(1)=\gamma\), then \[\begin{align} I'(s)&=\psi(s+1)+\gamma\\ I(s)&=\int \psi(s+1)\,ds +\gamma s+C\\ &=\int \frac{\partial}{\partial s}\bigg[\ln\Gamma(s+1)\bigg]\,ds +\gamma s+C\\ &=\ln\Gamma(s+1) +\gamma s+C\\ \end{align}\]

where \(\psi(z)\) is the digamma function. For \(s=0\), we have \(I(0)=0\). Implying \(C=0\), then

\[\begin{equation} I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s\qquad\qquad\square \end{equation}\]

For \(s=1\), we have

\[\begin{equation} I(1)=\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx=\gamma \end{equation}\]

where \(\gamma\) is The Euler–Mascheroni constant.

Problem 16 :

Prove

\[\begin{equation}\int_0^{\Large\frac\pi2}\cos^{n-1}x\cos ax\ dx=\frac{\pi}{2^n n\ \operatorname{B}\left(\frac{n+a+1}{2},\frac{n-a+1}{2}\right)} \end{equation}\]

where \(\operatorname{B}\left(x,y\right)\) is the beta function.

PS : POST YOUR SOLUTION BELOW EACH PROBLEM THREAD AND POST YOUR PROPOSED PROBLEM AS A NEW THREAD. PUT THEM IN SEPARATED THREAD. SO THAT THE POSTS LOOK MORE ORGANIZED. THANKS.

Solution 9: First, consider the following Lemma.

LEMMA : \[\int_{0}^{1} \frac{\ln{x}}{1-x^2} dx=-\frac{\pi^2}{8}\]

Proof: Since, \(\frac{1}{1-x^2}=\sum\limits_{n=0}^{\infty} {x^{2n}}\), we have,

\( \int_0^1 \frac{\ln{x}}{1-x^2}=\sum\limits_{n=0}^{\infty} \int_0^1 x^{2n}{\ln{x}} \: dx\)

\(=-\sum\limits_{n=0}^{\infty} {\frac{1}{(2n+1)^2}}\) (Using Integration By Parts)

\(=\sum\limits_{n=0}^{\infty} {\frac{1}{(2n)^2}} - \sum\limits_{n=0}^{\infty} {\frac{1}{(n)^2}}\)

\(=\frac{-3}{4}\zeta(2)= -\frac{\pi^2}{8}\)

Now,

\(\text{I}= \int_{0}^{1} \ln({\frac{1+x}{1-x}}) \frac{dx}{x\sqrt{1-x^2}}\)

Substituting \(x=\frac{1-y^2}{1+y^2}\), we have,

\(\text{I}=-4 \times \int_{0}^{1} \frac{\ln{y}}{1-y^2} dx\)

\(=\frac{\pi^2}{2}\) (Using the Lemma)

Solution of Problem 5 :

I will prove the general form of

\[\int_0^\infty\frac{x^{m-1}}{1+x^n}\,dx=\frac{\pi}{n\sin\frac{m\pi}{n}}\]

It can easily be proven by taking substitution \(\displaystyle y=\frac{1}{1+x^n}\) and the integral becomes Beta function

\[\frac{1}{n}\int_0^1 y^{\large 1-\frac{m}{n}-1}\ (1-y)^{\large \frac{m}{n}-1}\,dy=\frac{\Gamma\left(1-\frac{m}{n}\right)\Gamma\left(\frac{m}{n}\right)}{n}=\frac{\pi}{n\sin\frac{m\pi}{n}}\]

where the last part comes from Euler's reflection formula for the gamma function. Hence, by setting \(m=1\), we have

\[\int_0^\infty\frac{1}{1+x^n}\,dx=\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}=\frac{\pi}{n}\csc\left(\frac{\pi}{n}\right)\]

Problem 6 :

Show that

\[\int_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x}\,dx=\frac{\ln\left(3+2\sqrt{2}\right)}{2\sqrt{2}}\]

Ans:ln4a/2a^2