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# Brilliant Junior Integration Contest - Season 1

Hi Brilliant! Just like what Aditya Kumar and Anastasiya Romanova conducted earlier, this year we would also like to conduct an integration contest for juniors.

The aims of the Integration contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun.

Eligibility:- People should fulfill either of the 2 following

• 17 years or below

• Level 4 or below in Calculus

Eligible people here may participate in this contest.

The rules are as follows:

• I will start by posting the first problem. If there is a user solves it, then they must post a new one.

• You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

• Only make substantial comment that will contribute to the discussion.

• Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

• If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

• The scope of questions is only computation of integrals either definite or indefinite integrals.

• It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

• Do not copy questions from last year's contest. If anyone found to do so he/she will be banned from taking further part in this contest

• You are also NOT allowed to post a solution using a contour integration or residue method.

The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, digamma function,Harmonic numbers, trigonometric integral, Wallis' integral,

Format your post is as follows:

 1 2 3 4 5 6 7 **SOLUTION OF PROBLEM xxx (number of problem) :** **[Post your solution here]** **PROBLEM xxx (number of problem) :** **[Post your problem here]** 

Note by Rajdeep Dhingra
1 year, 7 months ago

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Problem 18:

Prove that: $\int_{0}^{1} \left(\frac{1}{\log x} + \frac{1}{1-x} -\frac{1}{2} \right) \frac{ \mathrm{d}x}{1-x} = -\frac{1}{2}+\frac{1}{2} \ln (2 \pi)-\frac{1}{2} \gamma.$

The Problem has been solved by Ishan Singh

- 1 year, 7 months ago

Lemma 1 : $$\displaystyle \sum_{r=1}^{n} H_{r} = (n+1)H_{n} - n$$

Proof : $$\displaystyle \sum_{r=1}^{n} H_{r} = \int_{0}^{1} \sum_{r=1}^{n} \dfrac{x^r-1}{x-1} \mathrm{d}x$$

$$\displaystyle = \int_{0}^{1} \dfrac{x^{n+1} - x - nx + n}{(x-1)^2} \mathrm{d}x$$

$$\displaystyle = \int_{0}^{1} \dfrac{x^{n+1} - x}{(x-1)^2} \mathrm{d}x - \int_{0}^{1} \dfrac{n}{x-1} \mathrm{d}x$$

$$\displaystyle = \text{A} + \text{B}$$

Using Integration By Parts on the first integral $$\text{A}$$, we have,

$$\displaystyle \text{A} = -\left[\dfrac{x^{n+1}-x}{x-1}\right]_{0}^{1} + \int_{0}^{1}\dfrac{(n+1)x^n - 1}{x-1} \mathrm{d}x$$

$$\displaystyle = -n + \int_{0}^{1}\dfrac{(n+1)x^n - 1}{x-1} \mathrm{d}x$$

$$\displaystyle \implies \text{A} +\text{B} = -n + (n+1) \cdot \int_{0}^{1}\dfrac{x^n - 1}{x-1} \mathrm{d}x$$

$$\displaystyle = (n+1)H_{n} - n$$

Lemma 2 : $$\displaystyle \int_{0}^{1} \dfrac{x^n - 1}{\ln x} \mathrm{d}x = \ln (n +1)$$

Proof : $$\displaystyle \int_{0}^{1} \dfrac{x^n - 1}{\ln x} \ \mathrm{d}x = \int_{0}^{1} \int_{0}^{n} x^y \ \mathrm{d}y \ \mathrm{d}x$$

$$\displaystyle = \int_{0}^{n} \int_{0}^{1} x^y \ \mathrm{d}x \ \mathrm{d}y$$

$$\displaystyle = \int_{0}^{n} \dfrac{1}{y+1} \ \mathrm{d}y$$

$$\displaystyle = \ln (n+1)$$

Lemma 3 : $$\displaystyle \int_{0}^{1} \left(\dfrac{1}{\ln (x)} + \dfrac{1}{1-x}\right) \mathrm{d}x = \gamma$$

Proof : $$\displaystyle \int_{0}^{1} \left( \dfrac{1}{1-x} + \dfrac{1}{\ln (x)}\right) \mathrm{d}x = \lim_{n \to \infty} \int_0^1\left(\dfrac{1-t^n}{1-t}-\frac{t^{n-1}-1}{\ln t} \mathrm{d}t \right)$$

$$\displaystyle = \lim _{n \to \infty} (H_{n} - \ln n)$$

$$\displaystyle = \gamma$$

Now,

$$\displaystyle \text{I} = \int_{0}^{1} \left(\dfrac{1}{\ln x} + \dfrac{1}{1-x} - \dfrac{1}{2}\right)\dfrac{\mathrm{d}x}{x-1}$$

$$\displaystyle = \lim_{n \to \infty} \sum_{r=1}^{n} \int_{0}^{1} \left(\dfrac{x^{r-1}}{\ln x} + \dfrac{x^{r-1}}{1-x} - \dfrac{x^{r-1}}{2} \right)$$

Using Lemma 2 and Lemma 3 , we have,

$$\displaystyle \text{I} = \lim_{n \to \infty} \sum_{r=1}^{n} \left(\gamma + \ln r - H_{r-1} -\dfrac{1}{2r} \right)$$

$$\displaystyle = \lim_{n \to \infty} \left( n\gamma + \ln (n!) - \sum_{r=1}^{n} H_{r-1} - \dfrac{H_{n}}{2} \right)$$

Using Lemma 1,

$$\displaystyle \text{I} = \lim_{n \to \infty} \left( n\gamma + \ln (n!) -n H_{n-1} + n - 1 -\dfrac{H_{n}}{2} \right)$$

Using Stirling's Approximation, we have,

$$\displaystyle \text{I} = \lim_{n \to \infty} \left( n\gamma + \dfrac{1}{2} \ln (2\pi) - n +\dfrac{1}{2} \ln (n) + n\ln (n) -n H_{n-1} + n - 1 -\dfrac{H_{n}}{2} \right)$$

Simplifying using $$\displaystyle \lim_{n \to \infty} (H_{n} - \ln n) = \gamma$$, we have,

$$\displaystyle \text{I} = \boxed{\dfrac{1}{2} \ln (2\pi) -\dfrac{1}{2} \gamma - \dfrac{1}{2}}$$

- 1 year, 7 months ago

Nice. I used Binet's formula.

- 1 year, 7 months ago

- 1 year, 7 months ago

@Aditya Kumar May post another question if he wants.

- 1 year, 7 months ago

wow,that was probably why i gave up halfway through lol

i'm sorry i didn't attend the integration contest until now,there was a problem with my wifi and it got fixed

- 1 year, 7 months ago

Problem 1

Find the value of the following $\Large \displaystyle \dfrac{\displaystyle \int_{-\infty}^{\infty}{{(x+1)}^2 e^{-{(x+1)}^2}dx}}{\displaystyle \int_{-\infty}^{\infty}{e^{-x^2}dx}}$

This problem has been solved by Aditya Kumar.

- 1 year, 7 months ago

Solution to Problem 1: ${ I }_{ 1 }=\int _{ -\infty }^{ \infty }{ { \left( x+1 \right) }^{ 2 }{ e }^{ { -\left( x+1 \right) }^{ 2 } } } dx$

Substitute: $${ \left( x+1 \right) }^{ 2 }=t$$

Therefore, we get: ${ I }_{ 1 }=\frac { 1 }{ 2 }\int _{ -\infty }^{ \infty }{ { \left( t \right) }^{ \frac { 1 }{ 2 } }{ e }^{ { - }t } } dt$

This is symmetric about $$x=0$$. Therefore, we can write it as: ${ I }_{ 1 }=\int _{ 0 }^{ \infty }{ { \left( t \right) }^{ \frac { 1 }{ 2 } }{ e }^{ { - }t } } dt$

${ I }_{ 1 }=\Gamma \left( \frac { 3 }{ 2 } \right) =\frac { 1 }{ 2 } \Gamma \left( \frac { 1 }{ 2 } \right) =\frac { \sqrt { \pi } }{ 2 }$

Similarly, $${ I }_{ 2 }=\sqrt { \pi }$$

Hence, $\boxed{\frac{{ I }_{ 1 }}{{ I }_{ 2 }}=\frac{1}{2}}$

- 1 year, 7 months ago

Problem 14:

If $$\displaystyle \int_{0}^{\infty} \dfrac{x^{3}}{1+e^{\alpha x}}dx = \dfrac{1}{64}$$
Find $$\alpha$$ given that , $$\alpha$$ is a positive constant and $$\zeta(4) = \dfrac{\pi^{4}}{90}$$

Problem has been solved by Rajdeep Dhingra and Aareyan Manzoor at the same instant. Aareyan has been given credits due to his solution.

- 1 year, 7 months ago

$I=\int_0^\infty x^3 \sum_{n=1}^\infty (-e^{-\alpha x})^n dx$ $I= \sum_{n=1}^\infty (-1)^n \int_0^\infty x^3 e^{-\alpha n x}dx=\sum_{n=1}^\infty (-1)^n \dfrac{6}{(\alpha n)^4}$ $I=\dfrac{6}{\alpha^4} (1-2^{-3}) \zeta(4)=\dfrac{1}{64}$ the result follows.

- 1 year, 7 months ago

Is the answer $$\huge \displaystyle \alpha = \pi {\left(\frac{56}{15}\right)}^{\frac14}$$ ?

- 1 year, 7 months ago

Problem 2:

Prove that: $\int _{ 0 }^{ 1 }{ \frac { \arctan { \left( \sqrt { { x }^{ 2 }+2 } \right) } }{ \sqrt { { x }^{ 2 }+2 } \left( { x }^{ 2 }+1 \right) } \ dx } =\frac { 5{ \pi }^{ 2 } }{ 96 }$

This problem has been solved by Harsh Shrivastava.

- 1 year, 7 months ago

Solution to Problem 2

$I = \int _{ 0 }^{ 1 }{ \frac { \arctan { \left( \sqrt { { x }^{ 2 }+2 } \right) } }{ \sqrt { { x }^{ 2 }+2 } \left( { x }^{ 2 }+1 \right) } \ dx } =\frac { 5{ \pi }^{ 2 } }{ 96 }$

Lets split $$I$$ as $$A \text{ and } B$$ by using the property that $$\arctan(x) = \pi /2 -\arctan(1/x)$$

Thus our integral becomes $$I = A-B$$,

where $$A = \displaystyle \pi /2\int_{0}^{1}\dfrac{dx}{(1+x^{2})(\sqrt{2+x^{2}})}$$

and $$B= \displaystyle \int_{0}^{1}\dfrac{\arctan(\frac{1}{\sqrt{x^{2} + 2}})}{(1+x^{2})(\sqrt{2+x^{2}})}$$

Evaluating A,

Let's evaluate A, without the limits,

$$\displaystyle \pi /2\int\dfrac{dx}{(1+x^{2})(\sqrt{2+x^{2}})} = \pi /2\arctan \dfrac{x}{\sqrt{x^{2} + 2}}$$

Which on substituting limits gives A = $$\pi ^{2}/12$$

Evaluating B,

Using $$1/a \arctan(1/a) = \int_{0}^{a} \frac{dy}{y^{2}+a^{2}}$$,

Let $$a = \sqrt{x^2+2}$$.

We get $$B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(2+x^{2} + y^{2})}$$

$$B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(1+y^{2})} - \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(2+x^{2} + y^{2})}$$

$$2B = \displaystyle \int_{0}^{1}\int_{0}^{1} \dfrac{dxdy}{(1+x^{2})(1+y^{2})}$$

$$2B = \displaystyle (\int_{0}^{1}\dfrac{dx}{1+x^{2}})^{2}$$

Thus $$B = \pi^{2} / 32$$

Hence $$I = A-B=\dfrac{5\pi^{2}}{96}$$

- 1 year, 7 months ago

- 1 year, 7 months ago

Problem 17

Prove that if $$c < 1$$, $\int_{0}^{\pi/2} {\arcsin{(c \cos{(x)})} \ dx} = \frac{c}{1^2} + \frac{c^3}{3^2} + \frac{c^5}{5^2} + \cdots \text{ (Upto Infinity)}$

The Problem has been solved by Aditya Kumar

- 1 year, 7 months ago

Solution to Problem 17:

Maclaurin series of $$\arcsin { \left( x \right) }$$ is: $$\displaystyle \arcsin { \left( x \right) } =\sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) }{ \sqrt { \pi } \left( 2n+1 \right) n! } { x }^{ 2n+1 } }$$

Here $$x=c \cos(x)$$

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) }{ \sqrt { \pi } \left( 2n+1 \right) n! } { \left( c\cos { \left( x \right) } \right) }^{ 2n+1 } } dx }$

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\sum _{ n=0 }^{ \infty }{ \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) { c }^{ 2n+1 } }{ \sqrt { \pi } \left( 2n+1 \right) n! } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \left( \cos { \left( x \right) } \right) }^{ 2n+1 }dx } }$

Therefore, by beta function we get:

$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \arcsin { \left( c\cos { \left( x \right) } \right) } dx } =\sum _{ n=0 }^{ \infty }{ \left( \frac { \Gamma \left( n+\frac { 1 }{ 2 } \right) { c }^{ 2n+1 } }{ \sqrt { \pi } \left( 2n+1 \right) n! }\frac { 1 }{ 2 } \frac { \Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( n+1 \right) }{ \Gamma \left( n+\frac { 3 }{ 2 } \right) } \right) }$

Hence, on simplifying, we get: $\boxed{ \displaystyle \int_{0}^{\pi/2} {\arcsin{(c \cos{(x)})} \ dx} = \frac{c}{1^2} + \frac{c^3}{3^2} + \frac{c^5}{5^2} + \cdots \text{ (Upto Infinity)}}$

- 1 year, 7 months ago

Problem 10

Let $S = \sin^2{(x)} + \dfrac{sin^4{(x)}}{3^2} + \dfrac{sin^6{(x)}}{5^2} + \cdots \text{ Upto Infinity }$

Prove that $\int_{0}^{\pi /2} {S \ dx} = \dfrac{{\pi}^2}{4} - \dfrac{\pi}2$

The Problem has been solved by Ishan Singh.

- 1 year, 7 months ago

I have come up with several ways to prove it. Here is one of the more interesting methods.

Lemma: $\displaystyle f(a) = \dfrac{1}{1+a} + \sum_{r=1}^{\infty} \left[ \dfrac{1}{(2r+a+1)}\prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \right] = \dfrac{\pi}{2^{a+1}} \dfrac{\Gamma (a+1)}{\Gamma^2\left(\dfrac{a+2}{2}\right)} \quad ; \quad a > -1$

Proof: I have proved it here

Now,

$$\displaystyle \text{I} = \int_{0}^{\pi /2} {S \ \mathrm{d}x} = \sum_{r=1}^{\infty}\int_{0}^1 \dfrac{\sin ^{2r}}{(2r-1)^2} \mathrm{d}x$$

$$\displaystyle = \dfrac{1}{2} \sum_{r=1}^{\infty} \dfrac{\operatorname{B}\left(r+\frac{1}{2} , \frac{1}{2} \right)}{(2r-1)^2}$$

$$\displaystyle = \dfrac{1}{2} \sum_{r=1}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right) \Gamma \left(\frac{1}{2} \right)}{(2r-1)^2 \Gamma(r+1)}$$

$$\displaystyle = \dfrac{\pi}{2} \sum_{r=1}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right) }{(2r-1)^2 \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}$$

Changing the index of summation and using $$\Gamma(x+1) = x\Gamma(x)$$, we have,

$$\displaystyle \text{I} = \dfrac{\pi}{4} \sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(r+1)(2r+1) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}$$

$$\displaystyle = \dfrac{\pi}{2} \left[\sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(2r+1) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)} - \sum_{r=0}^{\infty} \dfrac{\Gamma\left(r+\frac{1}{2} \right)}{(2r+2) \Gamma \left(\frac{1}{2}\right) \Gamma(r+1)}\right]$$

Using my solution here, we have,

$$\displaystyle \text{I} = \dfrac{\pi}{2} \left[ 1+ \sum_{r=1}^{\infty} \dfrac{1}{(2r+1)} \prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) - \dfrac{1}{2} - \sum_{r=1}^{\infty} \dfrac{1}{(2r+2)} \prod_{k=1}^{r}\left(\dfrac{2k-1}{2k}\right) \right]$$

$$\displaystyle = \dfrac{\pi}{2} \left[f(0) - f(1) \right]$$

$$\displaystyle = \boxed{\dfrac{\pi ^2}{4} - \dfrac{\pi}{2}}$$

- 1 year, 7 months ago

- 1 year, 7 months ago

More or less the same.

- 1 year, 7 months ago

@Ishan Singh @Rajdeep Dhingra Do any of you know another method to solve this question?

- 1 year, 7 months ago

Ishan wrote that he has come up with several ways so I guess has many. @Ishan Singh

- 1 year, 7 months ago

My other methods include using Dilogarithm and derivative of Beta function.

- 1 year, 7 months ago

Problem 6:

Find a closed form of the indefinite integral $\int H_{x} dx$

Where $$H_x$$ denotes harmonic number

This problem has been solved by Aditya Kumar.

- 1 year, 7 months ago

Solution to Problem 6:

I'll use the definition: $${ H }_{ x }=\psi \left( x+1 \right) +\gamma$$

$\int { { H }_{ x }dx } =\int { \left( \psi \left( x+1 \right) +\gamma \right) dx }$

We know that $$\displaystyle \psi \left( x \right) =\frac { d\left( x! \right) }{ dx }$$. Hence, using this, we get: $\boxed{\displaystyle \int { { H }_{ x }dx } =ln\left( x! \right) +\gamma x+c}$

- 1 year, 7 months ago

Problem 23

Prove that for an integer $$n\ge2$$

$$\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ n ]{ x } } \right\} } dx=\frac { n }{ n-1 } -\zeta (n)$$

hint:let $$x=\frac { 1 }{ u^{ n } }$$

- 1 year, 6 months ago

Let $$\displaystyle \text{I} = \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ \sqrt [ n ]{ x } } \right\} } dx$$

Substitute $$\displaystyle x = \frac{1}{t^n}$$

$$\displaystyle \implies \text{I} = n \int_{1}^{\infty} \dfrac{ \left \{t \right \} }{t^{n+1}} dt$$

$$\displaystyle = n \sum_{r=1}^{\infty} \int_{r}^{r+1} \dfrac{t-r}{t^{n+1}} dt$$

$$\displaystyle = n \sum_{r=1}^{\infty} \left[ \dfrac{t^{1-n}}{1-n} +r \dfrac{t^{-n}}{n} \right]_{r}^{r+1}$$

$$\displaystyle = -n \sum_{r=1}^{\infty} \left( \dfrac{1}{n(n+1) (r+1)^{n-1}} + \dfrac{1}{n(r+1)^n} \right)$$

$$\displaystyle = \dfrac{n}{n-1} - \zeta(n)$$

- 1 year, 6 months ago

Does the Curly brackets mean fractional part ?

- 1 year, 6 months ago

It is pointless to continue the contest. It hasn't garnered attention like the previous original integration contests.

- 1 year, 6 months ago

No Problem. Still it is open. Please post the next question if you want or if don't want to post, then @Hummus a will post.

- 1 year, 6 months ago

Note to the Participants:

• Please refrain from posting solution through images. It makes this note slow to load.

• Up-vote good solutions and problems. I have see that people are not up-voting at all. This is a wrong sign in the contest. Up-voting encourages the users to improve.

- 1 year, 7 months ago

Very True.I will change the note after it crosses 50 questions.

- 1 year, 7 months ago

Change it after 20. 50 is too much.

- 1 year, 7 months ago

Maybe you should do it now, my browser has hanged twice loading this note!

- 1 year, 7 months ago

Problem 9

Prove $\int_0^{\pi/2} x\ln(\tan(x)) dx= \dfrac{7\zeta(3)}{8}$

The Problem has been solved by Rajdeep Dhingra.

- 1 year, 7 months ago

Solution to Problem 9

Page 1

Page 2

- 1 year, 7 months ago

The second page is not at all clear. Please LaTeX your solution.

- 1 year, 7 months ago

Problem 27

Evaluate

$$\displaystyle\int _{0}^{1}{\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x+y } \right\} ^{ n }dxdy }}$$

with $$n\in\mathbb{N}$$

Hint:let $$x+y=t$$

- 1 year, 6 months ago

Problem 25

Let $$k$$ be a positive real number

then prove

$$\displaystyle\int _{0}^{1}{\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } =\frac { 2k+1 }{ (k+1)^{ 2 } } -\frac { \gamma }{ k+1 }}$$

Hint:let $$t=\frac{x^k}{y}$$

- 1 year, 6 months ago

Solution to Problem 25:

$I=\int _{ 0 }^{ 1 }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } }$

$I=\int _{ 0 }^{ 1 }{ \left( \int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dy } \right) dx }$

By IBP, we get: $I={ \left( \frac { { x }^{ k+1 } }{ k+1 } \int _{ { x }^{ k } }^{ \infty }{ \frac { \left\{ t \right\} }{ { t }^{ 2 } } dt } \right) }_{ 0 }^{ 1 }+\frac { k }{ k+1 } \int _{ 0 }^{ 1 }{ { x }^{ k }dx }$

Hence on simplifying, we get: $\displaystyle\int _{0}^{1}{\displaystyle\int _{ 0 }^{ 1 }{ \left\{ \frac { { x }^{ k } }{ y } \right\} dxdy } =\frac { 2k+1 }{ (k+1)^{ 2 } } -\frac { \gamma }{ k+1 }}$

- 1 year, 6 months ago

Problem 22 :

Prove That

$\int_{0}^{\infty} \dfrac{\sin \left(\frac{x^2 + 1}{2x} \right) \cos \left(\frac{x^2 - 1}{2x} \right) }{x} \mathrm{d}x = \dfrac{\pi}{2}$

This problem has been solved by Aditya Kumar.

- 1 year, 6 months ago

Solution to Problem 22:

Use the formula $$2\sin { \left( A \right) } \cos { \left( B \right) } =\sin { \left( A+B \right) } -\sin { \left( A-B \right) }$$

Therefore, we get: $I=\frac { 1 }{ 2 } \left\{ \int _{ 0 }^{ \infty }{ \frac { \sin { \left( x \right) } }{ x } } +\int _{ 0 }^{ \infty }{ \frac { \sin { \left( \frac { 1 }{ x } \right) } }{ x } dx } \right\}$

It is easy to see that both the integrals are same. Hence, $I=\int _{ 0 }^{ \infty }{ \frac { \sin { \left( x \right) } }{ x } } =\frac{\pi}{2}$

- 1 year, 6 months ago

@Rajdeep Dhingra post the next problem. I can't come up with one of this level.

- 1 year, 6 months ago

I don't have anymore problems. Any one can post the Next Question. @Samuel Jones@Harsh Shrivastava@Aareyan Manzoor

- 1 year, 6 months ago

Problem 8

Prove that $\int_{0}^{2 \pi} { e^{\cos{(\theta)}} \cos{(\sin{(\theta)})} \ d{\theta}} = 2\pi$

Not Original

The Problem has been solved by Aareyan Manzoor.

- 1 year, 7 months ago

solution to problem 8

Considering the function $$f(t) = \int_{0}^{2\pi} e^{t \cos(\theta)} \cos(t\sin(\theta)) \, d\theta$$, we see that $\begin{array}{rcl} tf'(t) & = & \displaystyle t \int_0^{2\pi} e^{t\cos\theta}\big[\cos\theta \cos(t\sin\theta) - \sin\theta\sin(t\sin\theta)\big]\,d\theta \\ & = & \displaystyle \int_0^{2\pi} \frac{\partial}{\partial \theta}\big[e^{t\cos\theta} \sin(t\sin\theta)\big]\,d\theta \\ & = & \Big[ e^{t\cos\theta}\sin(t\sin\theta)\Big]_0^{2\pi} \; = \; 0\\ &&f'(t)=0\to f(t)=C \end{array}$ Since $$f(0) = 2\pi$$, the desired integral is $$f(1) = \boxed{2\pi}$$.

- 1 year, 7 months ago

Problem 3:

Prove that $\displaystyle \int_{0}^{1} \dfrac{(\ln(1+x) - \ln(1-x))}{x(\sqrt{1-x^{2}})}dx=\frac{\pi^{2}}{2}$

This problem has been solved by Rajdeep Dhingra.

- 1 year, 7 months ago

Solution to Problem 3

Let us start with the integral $I(a) = \int_{0}^{1}{ \dfrac{\ln{(1+ax)}}{x\sqrt{1-x^2}} \ dx}$. Differentiate with respect to a. $I'(a) = \int_{0}^{1}{ \dfrac1{(1+ax)\sqrt{1-x^2}} \ dx}$ This can easily be solved by substituting $$1 + ax = \dfrac1t$$.We now get $I'(a) = \dfrac{\arccos{(a)}}{\sqrt{1-a^2}}$ Now we find I(a). $I(a) = \int {\dfrac{\arccos{(a)}}{\sqrt{1-a^2}} \ da} = -\dfrac12\arccos^2{(x)} + C$ Now we put a = 1. We get $I(1) = C$ Now we put a = -1. We get $I(-1) = -\dfrac{{\pi}^2}2 + C$ We get $I(1) - I(-1) = \dfrac{{\pi}^2}2$

Q.E.D

- 1 year, 7 months ago

Problem 26:

Prove that:

$\int_{0}^{\infty} \frac{\cos x - e^{-x^2}}{x} \ dx = -\frac{\gamma}{2}$

- 1 year, 6 months ago

Lemma : $\displaystyle \int_{0}^{\infty} \dfrac{\cos x - e^x}{x} \mathrm{d}x = 0$

Proof : $$\displaystyle \int_{0}^{\infty} \dfrac{\cos x - e^x}{x} \mathrm{d}x = \int_{0}^{\infty} \int_{0}^{\infty} \left(e^{-yx}\cos x - e^{-x(y+1)} \right) \mathrm{d}x \ \mathrm{d}y$$

$$\displaystyle = \int_{0}^{\infty} \left( \dfrac{y}{y^2 + 1} - \dfrac{1}{y+1} \right) \mathrm{d}y$$

$$\displaystyle = \left[\ln \left( \dfrac{\sqrt{y^2+1}}{y+1} \right)\right]_{y=0}^{y \to \infty} = 0$$

Proposition : $\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x = \gamma$

Proof : $$\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x$$

$$\displaystyle = \int_{0}^{\infty} \int_{0}^{1} (1-e^{-x})e^{-yx} \mathrm{d}x \ \mathrm{d}y - \int_{0}^{\infty} \int_{1}^{\infty} e^{-x(y+1)} \mathrm{d}x \ \mathrm{d}y$$

$$\displaystyle = \int_{0}^{\infty} \left(\dfrac{1-e^{-y}}{y} + \dfrac{e^{-(y+1)} - 1}{y+1} \right) \mathrm{d}y - \int_{0}^{\infty} \dfrac{e^{(y+1)}}{y+1} \mathrm{d}y$$

$$\displaystyle = -\int_{0}^{\infty} \left(e^{-y} - \dfrac{1}{y+1} \right)\dfrac{\mathrm{d}y}{y}$$

$$\displaystyle = -\int_{0}^{\infty} \left(e^{-x} - \dfrac{1}{x+1} \right)\dfrac{\mathrm{d}x}{x}$$

Using the Lemma,

$$\displaystyle = - \int_{0}^{\infty} \left(\cos x - \dfrac{1}{x+1} \right)\dfrac{\mathrm{d}x}{x}$$

$$\displaystyle = \gamma$$ (Using my solution here (see Problem 39))

Now,

$$\displaystyle \text{I} = \int_{0}^{\infty} \dfrac{\cos x - e^{-x^2}}{x} \mathrm{d}x$$

$$\displaystyle = \int_{0}^{\infty} \dfrac{e^{-x} - e^{-x^2}}{x} \mathrm{d}x \quad (\text{Using Lemma})$$

$$\displaystyle = - \int_{0}^{\infty} \dfrac{1-e^{-x}}{x} \mathrm{d}x + \int_{0}^{\infty} \dfrac{1-e^{-x^2}}{x} \mathrm{d}x$$

$$\displaystyle = -\int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty} \dfrac{1-e^{-x}}{x} \mathrm{d}x + \int_{0}^{1} \dfrac{1-e^{-x^2}}{x} \mathrm{d}x + \int_{1}^{\infty}\dfrac{1-e^{-x^2}}{x} \mathrm{d}x$$

$$\displaystyle = -\gamma + \int_{0}^{1} \dfrac{1-e^{-x^2}}{x} \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x^2}}{x} \mathrm{d}x \quad (\text{Using Proposition})$$

Substitute $$x^2 \mapsto x$$

$$\displaystyle \implies \text{I} = -\gamma + \dfrac{1}{2} \left( \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x}}{x} \mathrm{d}x \right)$$

$$\displaystyle = -\gamma + \dfrac{\gamma}{2} \quad (\text{Using Proposition})$$

$$\displaystyle = \boxed{-\dfrac{\gamma}{2}}$$

- 1 year, 6 months ago

Post the next problem.

- 1 year, 6 months ago

I don't have one appropriate for this contest right now, someone else may post the next problem.

- 1 year, 6 months ago

Nice one! There's a shorter method that uses rmt!

- 1 year, 6 months ago

R.M.T. on which function?

- 1 year, 6 months ago

RMT on both. You'll get $\lim _{ s\rightarrow 0 }{ \left( \Gamma \left( s \right) \cos \left( \frac { \pi \, s }{ 2 } \right) -\frac { 1 }{ 2 } \, \Gamma \left( \frac { s }{ 2 } \right) \right) } =\frac { -\gamma }{ 2 }$

- 1 year, 6 months ago

Nice!

- 1 year, 6 months ago

Problem 24:

Prove that

$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$

The Problem has been solved by Rajdeep Dhingra.

- 1 year, 6 months ago

Solution to Problem 24 :

Let $I = \int_{-\infty}^{\infty} { e^{- x^2} \ dx}$ We can write it as $I = \sqrt{I^2}$ $I = \sqrt{\int_{-\infty}^{\infty} { e^{- y^2} \ dy} \int_{-\infty}^{\infty} { e^{- x^2} \ dx}}$ $I = \sqrt{\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} { e^{- x^2 + y^2} \ dy \ dx }}$ Switching to Polar Coordinates of one integral. $I = \sqrt{\int_{0}^{2\pi} { \int_{0}^{\infty} { e^{-r^2} r \ dr \ d{\theta}}}}$ The 'dr' one integral can be calculated by integration by parts. The other one is very easy. $I = \sqrt{ \pi}$

- 1 year, 6 months ago

Nice! Please post the next problem.

- 1 year, 6 months ago

Problem 21:

Prove That:

$\int_{0}^1\dfrac {1}{x^x} dx = \sum_{n=1}^{\infty} \dfrac {1}{n^n}$

- 1 year, 6 months ago

Comment deleted Mar 28, 2016

solution to problem 21 $\int_0^1 x^{-x}dx=\int_0^1 e^{-x\ln(x)} dx= \int_0^1 \sum_{n=0}^\infty \dfrac{(-x\ln(x))^n}{n!} dx=\sum_{n=0}^\infty \dfrac{(-1)^n}{n!} \int_0^1 x^n\ln^n(x) dx$ Lemma :

$$\displaystyle \int_{0}^{1} x^{m}\ln^{n}(x)dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}}$$

Consider the integral,
$$\displaystyle \int_{0}^{1} x^{m}dx = \dfrac{1}{m+1}$$

Differentiate n times with respect to m,
$$\displaystyle \int_{0}^{1} x^{m}\ln^{n}x dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}}$$
put m=n to have our original integral $\sum_{n=0}^\infty \dfrac{(-1)^n}{n!} \int_0^1 x^n\ln^n(x) dx=\sum_{n=0}^\infty \dfrac{1}{(n+1)^{n+1}}=\sum_{n=1}^\infty \dfrac{1}{n^n}$

- 1 year, 6 months ago

Comment deleted Mar 31, 2016

Since @Aareyan Manzoor hasn't posted a problem, can I post (according to the rules) ?

- 1 year, 6 months ago

Yup.

- 1 year, 6 months ago

Posted!

- 1 year, 6 months ago

just a bit elaboration for your solution.

- 1 year, 6 months ago

You may post next problem, cause i am out of any good problem.

- 1 year, 6 months ago

Problem 20 prove $\int_0^1 \dfrac{\ln(x^2+x+1)}{x} dx=\dfrac{2\zeta(2)}{3}$

- 1 year, 6 months ago

$$\displaystyle \int_{0}^1 \dfrac{\ln (x^2+x+1)}{x} dx = \int_{0}^1 \dfrac{\ln (1-x^3)}{x} dx - \int_{0}^1 \dfrac{\ln (1-x)}{x} dx$$

$$\displaystyle = -\sum_{n=1}^{\infty} \int_{0}^1\dfrac {x^{3n-1}}{n} dx + \sum_{n=1}^{\infty} \int_{0}^1 \dfrac {x^{n-1}}{n} dx$$

$$\displaystyle = \dfrac {2}{3} \zeta (2)$$

- 1 year, 6 months ago

well done, you are correct.

- 1 year, 6 months ago

Awesome!

Post the next problem.

- 1 year, 6 months ago

Problem 11

Prove that $\int_{0}^{\pi/2} { {\left( \dfrac{x}{\sin{(x)}} \right)}^2 \ dx} = \pi \log{(2)}$

The Problem has been solved Aareyan Manzoor.

- 1 year, 7 months ago

solution to problem 11 $\int uv'=uv-\int u'v$ $I=\int_0^{\pi/2} x^2 \csc^2(x) dx$ $u=x^2\to u'=2x, v'=\csc^2 x\to v=-cot(x)$ $I=-x^2\cot(x)\mid_0^{\pi/2} +2\int_0^{\pi/2} x \cot(x) dx=2\int_0^{\pi/2} x \cot(x) dx$ $u=x\to u'=1, v'=\cot(x)\to v=-\ln(sin(x))$ $I=2x\ln(\sin(x))\mid_0^{\pi/2}+2\int_0^{\pi/2} \ln(\sin(x)) dx=\int_0^{\pi/2} \ln(\sin^2(x)) dx$ $B(a,1/2)=\int_0^{\pi/2} 2\sin^{2a-1}(x) dx$ $B'(a,1/2)=\int_0^{\pi/2} 2\ln(\sin^2(x))\sin^{2a-1}(x) dx$ $B'(a,1/2)=B(a,1/2)(\psi(a)-\psi(a+1/2))$ $a=1/2$ $\int_0^{\pi/2} 2\ln(\sin^2(x))dx=B(1/2,1/2)(\psi(1/2)-\psi(1))=\pi(-\gamma-(-\gamma-2\ln(2)))=2\pi\ln(2)$ $I=\int_0^{\pi/2} \ln(\sin^2(x)) dx=\boxed{\pi\ln(2)}$

- 1 year, 7 months ago

Problem 5:

Evaluate

$\displaystyle \int_{0}^{1} \dfrac{\ln (1+x)}{x}$

This problem was first solved by Aaryen Manzoor using a disallowed method and then solved by Aditya Kumar using a legal method. Credit is still given to Aaryen.

- 1 year, 7 months ago

Solution to Problem 5:

$I=\int _{ 0 }^{ 1 }{ \frac { \ln { \left( 1+x \right) } }{ x } dx }$

$I=\int _{ 0 }^{ 1 }{ \frac { \ln { \left( -x \right) } }{ 1-x } dx }$

$I=-\int _{ 0 }^{ 1 }{ \frac { \ln { \left( x \right) } }{ 1+x } dx }$

Now, I'll use power series of $$\frac{1}{x+1}$$, I'll write the integral as:

$I=-\sum _{ n=0 }^{ \infty }{ { \left( -1 \right) }^{ n }\int _{ 0 }^{ 1 }{ { x }^{ n }\ln { x } dx } }$

$I=\sum _{ n=0 }^{ \infty }{ \frac { { \left( -1 \right) }^{ n+1 } }{ { \left( n+1 \right) }^{ 2 } } }$

$\boxed{\therefore I=\frac { \zeta \left( 2 \right) }{ 2 } =\frac { { \pi }^{ 2 } }{ 12 } }$

- 1 year, 7 months ago

solution to problem 5

replace x with -x to have $\int_0^{-1} \dfrac{\ln(1-x)}{x} dx$ $\sum_{n≥1} \dfrac{x^n}{n}=\ln(1-x)$ dividing by x and integrating from 0 to a $\int_0^{a} \dfrac{\ln(1-x)}{x} dx=\sum_{n≥1} \dfrac{a^n}{n^2}$ We have $\int_0^{-1} \dfrac{\ln(1-x)}{x} dx= \sum_{n≥1}\dfrac{(-1)^n}{n^2}=\sum_{n≥1} \dfrac{(-1)^n}{n^2}=\sum_{n≥1} \dfrac{1}{n^2}-2\sum_{n≥1} \dfrac{1}{(2n)^2}=\\\dfrac{1}{2}\sum_{n≥1} \dfrac{1}{n^2}=\boxed{\dfrac{\pi^2}{12}}$

- 1 year, 7 months ago

Yeah there is a hell easy method.

:)

- 1 year, 7 months ago

$$\text{Alternate solution to Problem 5:}$$Using MacLaurin expansion of $$\ln(1+x)$$:$\ln(1+x)=\sum_{r=1}^{\infty} (-1)^{r+1}\dfrac{x^r}{r}$$\Longrightarrow \dfrac{\ln(1+x)}{x}=\sum_{r=1}^{r=\infty}(-1)^{r+1}\dfrac{x^{r-1}}{r}$$\Longrightarrow I=\int_{0}^{1}(\sum_{r=1}^{r=\infty}(-1)^{r+1}\dfrac{x^{r-1}}{r})dx$$\Longrightarrow I=\sum_{r=1}^{\infty}(-1)^{r+1}\dfrac{\color{red}{1}}{r^2}=\dfrac{{\pi}^2}{12}$.

- 1 year, 7 months ago

Problem 4

Prove $\int_{0}^{\infty} { \dfrac{\sqrt{x}}{(1+x)(2+x)(3+x)} \ dx} = \dfrac{\pi}2 \left(- 2\sqrt{2} - 1 + \sqrt{3}\right)$

This problem has been solved by Harsh Shrivastava.

- 1 year, 7 months ago

- 1 year, 7 months ago

hey @Rajdeep Dhingra i'm 16 yrs in age but level 5 in calculus , can i participate ?

- 11 months ago

Problem 19

Prove that $\int_0^1 \dfrac{\ln^2(1-x)}{x}dx=2\zeta(3)$

The Problem has been solved by Vighnesh Shenoy.

- 1 year, 7 months ago

Lemma :

$$\displaystyle \int_{0}^{1} x^{m}\ln^{n}(x)dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}}$$

Consider the integral,
$$\displaystyle \int_{0}^{1} x^{m}dx = \dfrac{1}{m+1}$$

Differentiate n times with respect to m,
$$\displaystyle \int_{0}^{1} x^{m}\ln^{n}x dx = \dfrac{(-1)^{n}n!}{(m+1)^{n+1}}$$

$$I = \displaystyle \int_{0}^{1} \dfrac{ln^{2}(1-x)}{x}dx = \int_{0}^{1} \dfrac{\ln^{2}(x)}{1-x} dx$$

Using the Taylor series for $$\dfrac{1}{1-x}$$

$$I = \displaystyle \sum_{k=0}^{\infty} \int_{0}^{1} x^{k} \ln^{2}(x)dx$$.
Using the Lemma,
$$I = \displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^{2} 2!}{(k+1)^{3}} = \sum_{k=1}^{\infty} \dfrac{2}{k^{3}} = 2\zeta(3)$$

- 1 year, 7 months ago

Post the next problem.

- 1 year, 6 months ago

Problem 16:

Find the closed form of the following indefinite integral,

$\displaystyle \int \dfrac{\arctan (x)}{x^{11}} dx$

This problem has been solved by Rajdeep Dhingra.

- 1 year, 7 months ago

Solution to Problem 16

It is a long tedious one. I will just mention some main steps.

Apply integration by parts with 1st function as $$\arctan{(x)}$$ and 2nd one as $$\frac{dx}{x^{11}}$$.

Now you will get an integral $$\displaystyle \int{\frac1{10x^{10}(x^2 + 1)} \ dx}$$.

Apply partial fractions you will get $\frac1{x^{10}} - \frac1{x^8} + \frac1{x^6} - \frac1{x^4} + \frac1{x^2} - \frac1{x^2 +1}$ Rest all is trivial Integration.

- 1 year, 7 months ago

Comment deleted Mar 23, 2016

Comment deleted Mar 23, 2016

Instead of partial fracs, you can substiture $$\dfrac{1}{x} = t$$

- 1 year, 7 months ago

Is the answer $$\displaystyle \large \sum_{k = 1}^{5}{\dfrac{{(-1)}^k}{10(2k-1){x}^{2k-1}}} - \frac{\arctan{(x)}}{10} - \frac{\arctan{(x)}}{10x^{10}}$$ ?

- 1 year, 7 months ago

Problem 13:

Evaluate $\displaystyle \int_{0}^{\pi /2} \ln (\cos x) dx$

The Problem has been solved by Rajdeep Dhingra and Vighnesh Shenoy. Vighnesh solved it late by 1second. Still credit is given to Vighnesh for the next Question.

- 1 year, 7 months ago

Solution to Problem 13

Let us start with $I = \int_{0}^{\pi/2} { \log{(\cos{(x)})} \ dx}$ Using properties we know that $I = \int_{0}^{\pi/2} { \log{(\sin{(x)})} \ dx}$ Adding both we get $2I = \int_{0}^{\pi/2} { \log{(\sin{(x)}\cos{(x)})} \ dx}$ Using little manipulation we get that $2I = \int_{0}^{\pi/2} { \log{(\sin{(2x)})} \ dx} - \int_{0}^{\pi/2} { \log{(2)} \ dx}$ Using lemma we get $2I = I - \dfrac{\pi}2 \log{(2)} \\ \Rightarrow \boxed{I = - \dfrac{\pi}2 \log{(2)} }$

Proof of Lemma Used

$I_1 = \int_{0}^{\pi/2} { \log{(\sin{(2x)})} \ dx}$ Substitute 2x = t. You will get $I_1 = \frac12 \int_{0}^{\pi} { \log{(\sin{(t)})} \ dt} \\I_1 = \frac12 \left( \int_{0}^{\pi/2} { \log{(\sin{(t)})} \ dt} + \int_{\pi/2}^{\pi} { \log{(\sin{(x)})} \ dx} \right)$ Now let us substitute $$x - \frac{\pi}2 = t$$ in the 2nd integral. Then using properties and manipulating. we get $I_1 = \int_{0}^{\pi/2} { \log{(\sin{(x)})} \ dx}$

- 1 year, 7 months ago

Comment deleted Mar 23, 2016

Comment deleted Mar 23, 2016

$$I =\displaystyle \int_{0}^{\dfrac{\pi}{2}} \ln(\cos x) dx$$
$$\therefore I =\displaystyle \int_{0}^{\dfrac{\pi}{2}} \ln(\sin x) dx$$

$$2I = I =\displaystyle \int_{0}^{\dfrac{\pi}{2}} \ln(\cos x \sin x) dx$$

$$2I =I =\displaystyle \int_{0}^{\dfrac{\pi}{2}} \ln(\sin 2x) dx - \int_{0}^{\dfrac{\pi}{2}} \ln(2)dx$$
In the first expression , substitute 2x = t .

$$2I = \displaystyle \dfrac{1}{2} \cdot \int_{0}^{\pi}{\ln(\sin t) \ dt} - \dfrac{\pi \ln(2)}{2}$$
$$2I =\displaystyle \dfrac{1}{2} \cdot 2 \cdot \int_{0}^{\dfrac{\pi}{2}} \ln(\sin t) dt - \dfrac{\pi \ln(2)}{2}$$
$$2I = I - \dfrac{\pi \ln(2)}{2}$$
$$I = \dfrac{-\pi \ln(2)}{2}$$

- 1 year, 7 months ago

Delete this question! This question is on brilliant. See this.

- 1 year, 7 months ago

Problem 12

$\int_0^1 \ln(x)\ln(1-x)x(1-x) dx$

Problem has been solved by Harsh Shrivastava.

- 1 year, 7 months ago

Simple derivative of beta function.

$$B(a,b) = \displaystyle \int_{0}^{1} x^{m-1} (1-x)^{n-1}dx$$

Required integral is $$\dfrac{\partial B(m,n)}{\partial m \partial n}$$ at $$m=n=2$$

This evaluates to$$\displaystyle \frac { \Gamma (m)\Gamma (n) }{ \Gamma (m+n) } (((\psi (m)-\psi (m+n))(\psi (n)-\psi (m+n))-\psi '(m+n))$$.

On putting the values of m and n, the final answer evaluates to $$0.0684$$.

- 1 year, 7 months ago

This is not a complete answer. Please show how you got 0.0684.

- 1 year, 7 months ago

Problem 7:

Prove that $\int_0^1 \log \Gamma(x)\cos (16\pi x)\, dx=\frac{1}{32}$

This problem has been solved by Rajdeep Dhingra.

- 1 year, 7 months ago

Let us start with $I = \int_{0}^{1}{ \log{(\Gamma{(x)})}cos(16\pi x) \ dx}$ We can also write $I = \int_{0}^{1}{ \log{(\Gamma{(1-x)})}cos(16\pi - 16\pi x) \ dx} = \int_{0}^{1}{ \log{(\Gamma{(1-x)})}cos(16\pi x) \ dx}$ Adding both we get $2I = \int_{0}^{1}{ \log{(\Gamma{(1-x)}\Gamma{(x)})}cos(16\pi x) \ dx}$ Using Euler Reflection Formula we get $2I = \int_{0}^{1}{ \log{(\pi)}cos(16\pi x) \ dx - \int_{0}^{1}{ \log{(\sin{(\pi x)})}cos(16\pi x) \ dx}}$ Both are standard computations. $2I = 0 - \left(-\dfrac1{16}\right)$ $\boxed{I = \frac{1}{32}}$ Q.E.D

- 1 year, 7 months ago

Problem 15:

Time for an indefinite integral:

$\int \dfrac{\sin^9 x}{\cos^8 x} dx$

This problem has been solved by Vighnesh Shenoy and Aditya Kumar at almost the same time. Credits has been given to Vighnesh.

- 1 year, 7 months ago

Solution to Problem 15:

$I=\int { \frac { \sin ^{ 9 }{ x } }{ \cos ^{ 8 }{ x } } dx }$

Substitute: $$\cos x=t$$

$I=-\int { \left( { t }^{ -8 }-{ 4t }^{ -6 }+{ 6t }^{ -4 }-{ 4t }^{ -2 }+1 \right) } dt$

$\therefore I=-\cos { \left( x \right) } +\frac { \sec ^{ 7 }{ \left( x \right) } }{ 7 } -\frac { 4\sec ^{ 5 }{ \left( x \right) } }{ 5 } +2\sec ^{ 3 }{ \left( x \right) } -4\sec { \left( x \right) }$

- 1 year, 7 months ago

$$\cos x = t$$
$$\therefore -\sin x dx = dt$$
$$I = \displaystyle \int \dfrac{(1-t^{2})^{4}}{t^{8}}\cdot -dt$$
$$I =- \displaystyle \int \dfrac{1-4t^{2} + 6t^{4} -4t^{6} + t^{8}}{t^{8}} dt$$
$$I = - \left( \dfrac{-1}{7t^{7}} + \dfrac{4}{5t^{5}} - \dfrac{6}{3t^{3}} + \dfrac{4}{t} + t \right)$$ + c $$I = - \left( \dfrac{-1}{7\cos^{7} x} + \dfrac{4}{5\cos^{5} x} - \dfrac{6}{3\cos^{3}x}+ \dfrac{4}{\cos x} + \cos x \right) + c$$

- 1 year, 7 months ago

Lol I was typing. So I didn't see. Post the next problem.

- 1 year, 7 months ago

Oh no! I was about to start typing...

- 1 year, 7 months ago