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

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

Now consider

$I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt\qquad a\ge0$

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

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

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

$I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt=\frac{\pi}{2}-\arctan(a)$

and

$I(0)=\int_0^\infty\frac{\sin t}{t}\,dt=\frac{\pi}{2}$

Thus

$\int_0^\infty\frac{\sin^2x}{x^2}\,dx=\frac{\pi}{2}$

Problem 4 :

Prove

$\int_0^{\pi/2}\frac{\ln(\cos x)}{\sin x}\,dx=-\frac{\pi^2}{8}$

Solution of Problem 7:

Proposition :

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

Proof :

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

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

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

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

Problem 8 :

Prove

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

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

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

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

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

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

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

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

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

Problem 12 :

Prove

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

$Solution\quad of\quad Problem\quad 12$

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 $\int_0^1 \left(\frac{1}{1-x} + \frac{1}{\ln x} \right)\,dx=\int_0^1 \frac{\ln x+1-x}{(1-x)\ln x}\,dx$

Proposition :

$I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx=\ln\Gamma(s+1) +\gamma s$

Proof :

Let

$I(s)=\int_0^1 \frac{s\ln x+1-x^s}{(1-x)\ln x}\,dx$

then

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