Brilliant Integration Contest - Season 1

I am interested in holding an Integration Contest here on Brilliant.org like any other online forums such as AoPS or Integrals and Series. 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. Anyone here may participate in this contest.

The rules are as follows

  1. I will start by posting the first problem. If there is a user solves it, then (s)he must post a new one.
  2. 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.
  3. Please make a substantial comment.
  4. Make sure you know how to solve your own problem before posting it in case there is no one can answer it within a week, then you must post the solution and you have a right to post another problem.
  5. 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.
  6. The scope of questions is only computation of integrals either definite or indefinite integrals.
  7. You are NOT allowed to post a multiple integrals problem as well as a complex integral problem.
  8. You are also NOT allowed to post a solution using a contour integration or residue method.
  9. The final answer can ONLY contain the following special functions: gamma function, beta function, Riemann zeta function, Dirichlet eta function, dilogarithm, digamma function, and trigonometric integral.

Please post your solution and your proposed problem in a single new thread.

Format your post is as follows:

SOLUTION OF PROBLEM xxx (number of problem) :

[Post your solution here]

PROBLEM xxx (number of problem) :

[Post your problem here]

Please share this note so that lots of users here know this contest and take part in it. (>‿◠)✌

Okay, let the contest begin! Here is the first problem:

PROBLEM 1 :

For a>0a>0, show that

0alnxaxx2dx=πln(a4) \int_0^a \frac{\ln x}{\sqrt{ax-x^2}}\,dx=\pi\ln\left(\frac{a}{4}\right)

P.S. You may also want to see Brilliant Integration Contest - Season 1 (Part 2) and Brilliant Integration Contest - Season 1 (Part 3).

Note by Anastasiya Romanova
4 years, 7 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

Here is the solution: I=0aln xaxx2dx\displaystyle I = \int \limits_0^a \frac{ \text{ln } x }{\sqrt{ax - x^2}} \text{d}x

I=0aln (ax)axx2dx\displaystyle I = \int \limits_0^a \frac{ \text{ln } (a-x) }{\sqrt{ax - x^2}} \text{d}x

2I=0aln (axx2)axx2dx\displaystyle \Rightarrow 2I = \int \limits_0^a \frac{ \text{ln } (ax - x^2) }{\sqrt{ax - x^2}} \text{d}x

Put x=a2(1sint)axx2=a24cos2t\displaystyle x = \frac{a}{2} (1 - \sin t ) \Rightarrow ax - x^2 = \frac{a^2}{4} \cos ^2 t

Therefore, the integral becomes, 2I=π2π2ln (a24cos2t)a24cos2t(a2cost)(dt)\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,
2I=40π2ln (a2cost)dt\displaystyle 2I = 4\int \limits_0^{\frac{\pi}{2}} \text{ln } (\frac{a}{2} \cos t ) \text{d}t

The value of 0π2ln (cosθ)dθ\displaystyle \int \limits_0^{\frac{\pi}{2}} \text{ln } ( \cos \theta ) \text{d}\theta is π ln 22\displaystyle \frac{ -\pi \text{ ln } 2 }{2} (which I have calculated separately and I can post if it is required).

Thus, we obtain, I=π ln (a2)2π ln 22=π ln (a4)\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)

Sudeep Salgia - 4 years, 7 months ago

Log in to reply

Find a closed form expression for the integral : I=sin(2015x)sin2013x dx\displaystyle I = \int \sin (2015 x) \sin ^{2013} x \text{ d}x

Sudeep Salgia - 4 years, 7 months ago

Log in to reply

Solutionofproblem2 Solution \quad of \quad problem \quad 2

Split sin(2015x)sin(2015x) as sin(2014x+x)sin(2014x+x) and our integral becomes :

I=(sin2014x.cos(2014x)+sin(2014x)cos(x)sin2013(x))dxI=\displaystyle \int { ({ sin }^{ 2014 }x.cos(2014x)+sin(2014x)cos(x){ sin }^{ 2013 }(x))\quad dx }

Multiply and divide it by 20142014 to get :

12014(sin2014x.(2014cos(2014x))+sin(2014x)(2014sin2013(x)cos(x)))dx\frac { 1 }{ 2014 } \displaystyle \int { ({ sin }^{ 2014 }x.(2014cos(2014x))+sin(2014x)(2014{ sin }^{ 2013 }(x)cos(x)))\quad dx }

I=12014sin2014(x)dsin(2014x)+sin(2014x)(dsin2014x)\Rightarrow \displaystyle I=\frac { 1 }{ 2014 } \int { { sin }^{ 2014 }(x)dsin(2014x)+sin(2014x)(d{ sin }^{ 2014 }x) }

I=12014d(sin2014(x)sin(2014x))\Rightarrow \displaystyle I=\frac { 1 }{ 2014 } \int { d({ sin }^{ 2014 }(x)sin(2014x)) }

I=sin2014(x)sin(2014x)2014+C\large \displaystyle I=\frac { { sin }^{ 2014 }(x)sin(2014x) }{ 2014 } +C

Problem3\large Problem \quad 3

Evaluate I=0(sin(x)x)2dxI=\displaystyle \int _{ 0 }^{ \infty }{ { (\frac { sin(x) }{ x } })^{ 2 }dx }

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

@Ronak Agarwal Perfect. Nice work Ronak. Expecting a problem soon.

Sudeep Salgia - 4 years, 7 months ago

Log in to reply

@Ronak Agarwal Nicely done, +1. Now you must post a new problem, but before that. I think it would be nice if you post this as a new post, no need to reply @Sudeep Salgia. See the rules & the format post. Thank you. :)

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

@Ronak Agarwal Where is your proposed problem? We are waiting it. Please don't let us wait too long like this. You should post your solution and your proposed problem at once.

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

@Anastasiya Romanova @Anastasiya Romanova I have posted my problem sorry for holding the contest.

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

@Ronak Agarwal It's okay. I have posted the solution of your problem and also proposed a new problem. :)

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

Nicely done, +1! But you forget to post your problem anyway. You have a privilege for that so that this contest keeps going on. Post PROBLEM 2 in your thread solution. Thanks :)

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

Well, there must be a rule for people like Ronak! Solving a problem doesn't give you the right to withhold the contest. In such a case, someone should be allowed to interfere, right?

Danny Kills - 4 years, 7 months ago

Log in to reply

@Danny Kills Sorry If that troubled all because actually I have to go to coaching centre, and my internet connection got down and I went to my coaching centre, I have returned just now and has posted the question.

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

@Danny Kills OK, if no-one posts his/her own problem after solving a problem within a day. The one who has a right to post a problem is the last solver before him/her. Is this fair?

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

Solution of Problem 3 :

Use integration by parts by taking u=sin2xu=\sin^2x and dv=dxx2dv=\dfrac{dx}{x^2}, we get

0sin2xx2dx=sin2xx0+02sinxcosxxdx=0+0sin2xxdx=0sinttdtt=2x\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)=0eatsinttdta0 I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt\qquad a\ge0

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

I(a)=0eatsintdt=11+a2integration by parts twiceI(a)=11+a2da=arctan(a)+C\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()=0I(\infty)=0, implying C=π2C=\dfrac{\pi}{2}. Hence

I(a)=0eatsinttdt=π2arctan(a) I(a)=\int_0^\infty\frac{e^{-at}\sin t}{t}\,dt=\frac{\pi}{2}-\arctan(a)

and

I(0)=0sinttdt=π2 I(0)=\int_0^\infty\frac{\sin t}{t}\,dt=\frac{\pi}{2}

Thus

0sin2xx2dx=π2 \int_0^\infty\frac{\sin^2x}{x^2}\,dx=\frac{\pi}{2}

Problem 4 :

Prove

0π/2ln(cosx)sinxdx=π28 \int_0^{\pi/2}\frac{\ln(\cos x)}{\sin x}\,dx=-\frac{\pi^2}{8}

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

SolutionofProblem4\large Solution\quad of\quad Problem\quad 4

Firstly we will prove a general result

Result

01xaln(x)dx=1(1+a)2\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a }ln(x)dx } =\frac { -1 }{ { (1+a) }^{ 2 } }

Proof

I=01xaln(x)dxI=\displaystyle \int _{ 0 }^{ 1 }{ { x }^{ a }ln(x)dx }

Integrating by parts u=ln(x),dv=xadxu=ln(x),dv={x}^{a}dx

I=ln(x)xa+1a+1011a+101xadxI=\displaystyle ln(x)\frac { { x }^{ a+1 } }{ a+1 } \overset { 1 }{ \underset { 0 }{ | } } -\frac { 1 }{ a+1 } \int _{ 0 }^{ 1 }{ { x }^{ a }dx }

I=1(1+a)2I=\frac { -1 }{ { (1+a) }^{ 2 } }

Hence proved

Now we have :

I=0π2ln(cos(x))sin(x)dxI=\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { ln(cos(x)) }{ sin(x) } dx }

Take cos(x)=ycos(x)=y to get our integral as :

01ln(y)dy1y2\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(y)dy }{ 1-{ y }^{ 2 } } }

Now 11y2=n=0y2n\frac { 1 }{ 1-{ y }^{ 2 } } =\displaystyle \sum _{ n=0 }^{ \infty }{ { y }^{ 2n } }

Our integral becomes :

I=n=001y2nln(y)dyI=\displaystyle \sum _{ n=0 }^{ \infty }{ \int _{ 0 }^{ 1 }{ { y }^{ 2n }ln(y)dy } }

Using our proved result we get :

I=n=01(2n+1)2I=-\displaystyle \sum _{ n=0 }^{ \infty }{ \frac { 1 }{ { (2n+1) }^{ 2 } } }

Also the summation can we written as :

I=(n=11n2n=11(2n)2)I=-(\displaystyle \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } } -\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { (2n) }^{ 2 } } } )

I=(ζ(2)14ζ(2))=34ζ(2)=π28I=-(\zeta (2)-\frac { 1 }{ 4 } \zeta (2))=\frac { -3 }{ 4 } \zeta (2)=\large \frac{-{\pi}^{2}}{8}

Where ζ(x)\zeta(x) is the zeta function

Problem5\large Problem \quad 5

Find closed form of I=0dx1+xn\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ n } } }

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

Keep going fella, +1! But please post your solution & proposed problem in a new thread like I did. No need to reply other threads to post a solution & a problem. Anyway, I've posted the solution of your problem & also proposed a new problem.

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

Solution of Problem 7:

Proposition :

0πdxp+cosxdx=πp21\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(x2)t=\tan\left(\dfrac{x}{2}\right), then

0πdxp+cosxdx=02p+1+(p1)t2dtt=p+1p1tany=2p21arctant  0=πp21Q.E.D.\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. pp twice and setting p=10p=\sqrt{10}, we have

2p20π1p+cosxdx=2p2[πp21]0π2(p+cosx)3dx=π(2p2+1)(p21)50π1(10+cosx)3dx=7π162\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

0π2dx1+8sin2(tanx)=π6(2e2+12e21) \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)

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

Solution of Problem 8:

Substitute tanxx\tan x\mapsto x, then the integral is:

0dx(1+x2)(1+8sin2x)=0dx(1+x2)(54cos(2x))\int_0^{\infty} \frac{dx}{(1+x^2)(1+8\sin^2x)}=\int_0^{\infty} \frac{dx}{(1+x^2)(5-4\cos(2x))}

=011+x2(13(1+2k=1cos(2kx)2k))=π6+23k=112k0cos(2kx)1+x2dx=\int_0^{\infty} \frac{1}{1+x^2}\left(\frac{1}{3}\left(1+2\sum_{k=1}^{\infty} \frac{\cos(2kx)}{2^k}\right)\right)=\frac{\pi}{6}+\frac{2}{3}\sum_{k=1}^{\infty} \frac{1}{2^k}\int_0^{\infty} \frac{\cos(2kx)}{1+x^2}\,dx

=π6+π3k=1(12e2)k=π6+π312e21=π6(2e2+12e21)=\frac{\pi}{6}+\frac{\pi}{3}\sum_{k=1}^{\infty} \left(\frac{1}{2e^2}\right)^k=\frac{\pi}{6}+\frac{\pi}{3}\frac{1}{2e^2-1}=\boxed{\dfrac{\pi}{6}\left(\dfrac{2e^2+1}{2e^2-1}\right)}

I have used the following result:

0cos(mx)x2+a2dx=π2aeam\int_0^{\infty} \frac{\cos(mx)}{x^2+a^2}\,dx=\frac{\pi}{2a}e^{-am}

I cannot think of a challenging problem at the moment, I request somebody else to post one. Thanks!

Pranav Arora - 4 years, 7 months ago

Log in to reply

Ingenious! But I have a doubt - In your fourth step, how is it guaranteed that the function cos(2kx)2k\dfrac{\cos (2kx)}{2^k} is positive on (0,)(0,\infty)? PS : you forgot to write the dx\mathrm{d}x in the third step.

Pratik Shastri - 4 years, 7 months ago

Log in to reply

@Pratik Shastri I don't know what did you mean by positive/negative on (0,)(0,\infty), but if you meant to swap between integral & summation sign, it is valid because cos(2kx)1+x2\frac{\cos(2kx)}{1+x^2} is continuous, finite & integrable, therefore swapping those two signs can be justified by Fubini's theorem.

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

@Anastasiya Romanova If pranav refuses to put up a question then who put will put it up @Anastasiya Romanova .

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

@Ronak Agarwal According to the rule 4, the one is the last solver and it turns out that one is me. See PROBLEM 9 below, I've just proposed it.

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

@Anastasiya Romanova Right. I confused it with Tonelli's theorem.

Pratik Shastri - 4 years, 7 months ago

Log in to reply

Sorry for very late response. Using elementary techniques,-

Solution - Problem 7

0πdxpcosx \displaystyle \int_{0}^{\pi} \dfrac{dx}{p - cosx}

2I=0π2p dxp2cos2x 2I = \displaystyle \int_{0}^{\pi} \dfrac{2p~dx}{p^2 - cos^2x}

I=0πp dxp2(1+tan2x)1 I = \displaystyle \int_{0}^{\pi} \dfrac{p~dx}{p^2(1 + tan^2x) - 1}

I=p0πsec2x dxp2tan2x+(p21)2 I = p \displaystyle \int_{0}^{\pi} \dfrac{sec^2x~dx}{p^2tan^2x + \sqrt{(p^2 - 1)^2}}

tanx = t

I=2p0dtp2t2+(p21)2 I = 2p \displaystyle \int_{0}^{\infty} \dfrac{dt}{p^2t^2 + \sqrt{(p^2 - 1)^2}}

I=2pp21×1p[0tan1ptp21 I = \dfrac{2p}{ \sqrt{p^2 - 1}} \times \dfrac{1}{p} \Big[_{0}^{\infty} tan^{-1} \dfrac{pt}{ \sqrt{p^2 - 1}}

I=πp21 I = \dfrac{\pi}{ \sqrt{p^2 - 1}}

U Z - 4 years, 5 months ago

Log in to reply

Solution to problem 10

First integrate by parts -

I=0sin3xx2dx=03sin2xcosxxdxI= \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 L{f(t)t}(s)=sF(p)dp\mathcal{L} \left\{\dfrac{f(t)}{t}\right\}(s)=\displaystyle\int_{s}^{\infty} F(p) dp (in our case, s0s \rightarrow 0).

I=30L{sin2tcost}(p) dpI=3\int_{0}^{\infty} \mathcal{L} \{\sin^2{t} \cos{t}\}(p) \ dp

I=302s(s2+1)(s2+9)ds=3log34I=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 : L{f(t)}(s)=0estf(t)dt\mathcal{L}\{f(t)\}(s)=\displaystyle\int_{0}^{\infty} e^{-st} f(t) \mathrm{d}t

Problem 11

Find

0logxx2+α2dx     for α>0\int_{0}^{\infty} \dfrac{\log {x}}{x^2+\alpha^2} \mathrm{d}x \ \ \ \ \ \text{for} \ \alpha>0

Pratik Shastri - 4 years, 7 months ago

Log in to reply

Solution of Problem 13 :

First we prove the following proposition:

Proposition :

01xalnkx dx=(1)kk!(a+1)k+1for  k=0,1,2,\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 01xa dx=1a+1for  α>1. \int_0^1 x^a\ dx=\frac1{a+1}\qquad\text{for }\ \alpha>-1. Differentiating equation above kk times w.r.t. aa we have 01kak(xa) dx=01xalnkx dx=(1)kk!(a+1)k+1Q.E.D. \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

0xa1ebx1dx=0xa1ebx1ebxdxfor  a,b>0\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=ebxy=e^{-bx}, then

0xa1ebx1dx=(1)a1ba01lna1y1ydy\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 11y\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

0xa1ebx1dx=(1)a1ba01k=0yklna1ydy=(1)a1bak=001yklna1ydy=(1)a1bak=0(1)a1(a1)!(k+1)a=(a1)!bak=11ka=Γ(a)ζ(a)ba\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 Γ(a)\Gamma(a) is the gamma function and ζ(a)\zeta(a) is the Riemann zeta function.

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

0xnex1dx=Γ(n+1)ζ(n+1)\int_0^\infty\frac{x^{n}}{e^{x}-1}\,dx=\Gamma(n+1)\,\zeta(n+1)

Problem 14 :

Prove that

01ln(3+x3x)dxx(1x)=πln(7+435+26)\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}

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

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

01ln(1+x1x)dxx1x2=π22\int_0^1 \ln\left(\frac{1+x}{1-x}\right)\frac{dx}{x\sqrt{1-x^2}}=\frac{\pi^2}{2}

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

Solution -Problem - 9 We substitute, t=1+x1x    x=1t1+t t = \frac{1+x}{1-x} \implies x = \frac{1-t}{1+t} Doing the substitution and , simplifying , gives, I=01ln(t)(t)(1t)dt I = - \int_{0}^{1} \frac{\ln(t)}{(\sqrt{t})(1-t)} dt Now we substitute, t=sin2(x) t=\sin^2(x) I=40π2ln(sin(x))cos(x)dx I = -4\int_{0}^{\frac{\pi}{2}} \frac{\ln(\sin(x))}{\cos(x)} dx Using, Problem -44 I=4×π28=π22 I = -4 \times \frac{-\pi^2}{8} = \frac{\pi^2}{2}

Shivang Jindal - 4 years, 7 months ago

Log in to reply

I had already posted the solution to the problem and a new problem also you should not reply instead post your solution as a seperate comment as I did @Shivang Jindal

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

@Ronak Agarwal Lol, time difference of 5mins. I think i was typing solution when you posted the solution .

Shivang Jindal - 4 years, 7 months ago

Log in to reply

@Shivang Jindal You can try my posted question. @Shivang Jindal

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

I am able to write the integral as a series - I=πn=0(2n)!22n(n!)2(2n+1)I=\pi \sum_{n=0}^{\infty} \dfrac{(2n)!}{2^{2n}(n!)^2(2n+1)}

Pratik Shastri - 4 years, 7 months ago

Log in to reply

Solutionofproblem9Solution\quad of\quad problem\quad 9

I=01ln(1+x1x)dxx1x2\displaystyle I=\int _{ 0 }^{ 1 }{ ln(\frac { 1+x }{ 1-x } )\frac { dx }{ x\sqrt { 1-{ x }^{ 2 } } } }

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

0π2ln(cot2θ2)dθcosθ\displaystyle \int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln({ cot }^{ 2 }\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } }

I=(2)0π2ln(tanθ2)dθcosθ\Rightarrow \displaystyle I=(-2)\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(tan\frac { \theta }{ 2 } )\frac { d\theta }{ cos\theta } }

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

(4)01ln(t)dt1t2\displaystyle (-4)\int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } }

I proved in my solution to problem 4 that :

01ln(t)dt1t2=π28\displaystyle \int _{ 0 }^{ 1 }{ \frac { ln(t)dt }{ 1-{ t }^{ 2 } } } =\frac { -{ \pi }^{ 2 } }{ 8 }

Using this I get :

I=π22I=\frac{{\pi}^{2}}{2}

Problem10Problem\quad 10

Find 0sin3xx2dx\displaystyle \int _{ 0 }^{ \infty }{ \frac { { sin }^{ 3 }x }{ { x }^{ 2 } } dx }

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

Solution of Problem 11 :

Using substitution u=α2x    x=α2u    dx=α2u2 duu=\dfrac{\alpha^2}{x}\;\Rightarrow\;x=\dfrac{\alpha^2}{u}\;\Rightarrow\;dx=-\dfrac{\alpha^2}{u^2}\ du, then

0lnxx2+α2 dx=0ln(α2u)(α2u)2+α2α2u2 duI(α)=02lnαlnuα2+u2 du=2lnα01α2+u2 du0lnuu2+α2 du=2lnα01α2+u2 duI(α)I(α)=lnα01α2+u2 du.\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=αtanθu=\alpha\tan\theta, then

0lnxx2+α2 dx=lnαα0π2 dθ=πlnα2α\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

cos(arctan2x)(1+x2)1+4x2dx=π3 \int_{-\infty}^{\infty}\frac{\cos \left(\, \arctan 2x\right)}{(1+x^2)\sqrt{1+4x^2}}\,dx=\frac{\pi}{3}

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

The last substitution should be u=αtanθu=\alpha \tan \theta :)

Pratik Shastri - 4 years, 7 months ago

Log in to reply

Fixed it. Sorry, I'm too hasty. Thanks... :)

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

This is a simple kind of NCERT problem. It can be easily done by substituting t=tan1xα t = tan^{-1}\dfrac{x}{\alpha} and then applying by parts

U Z - 4 years, 5 months ago

Log in to reply

SolutionofProblem12Solution\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 :

I=20cos(arctan(2x))(1+x2)1+4x2dx\displaystyle I=2\int _{ 0 }^{ \infty }{ \frac { cos(arctan(2x)) }{ (1+{ x }^{ 2 })\sqrt { 1+4{ x }^{ 2 } } } dx }

Now we know that cos(arctan(2x))=11+4x2\displaystyle cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}. Using this our integral becomes :

I=201(1+x2)(1+4x2)dx\displaystyle I=2\int _{ 0 }^{ \infty }{ \frac { 1 }{ (1+{ x }^{ 2 })(1+4{ x }^{ 2 }) } dx }

Splliting it by partial fractions we get :

I=23(0dx(14+x2)0dx1+x2)\displaystyle I=\frac { 2 }{ 3 } (\int _{ 0 }^{ \infty }{ \frac { dx }{ (\frac { 1 }{ 4 } +{ x }^{ 2 }) } } -\int _{ 0 }^{ \infty }{ \frac { dx }{ 1+{ x }^{ 2 } } } )

I=23(2tan1(2x)0tan1(x)0)\displaystyle I=\frac { 2 }{ 3 } (2{ tan }^{ -1 }(2x)\overset { \infty }{ \underset { 0 }{ | } } -{ tan }^{ -1 }(x)\overset { \infty }{ \underset { 0 }{ | } } )

I=π3\displaystyle I=\frac { \pi }{ 3 }

Problem13Problem\quad 13

Find closed form of I=0xnex1dx\displaystyle I=\int _{ 0 }^{ \infty }{ \frac { { x }^{ n } }{ { e }^{ x }-1 } dx }

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

Wait a sec!? Could you rectify your solution, because the original problem is written as arctan(2x)\arctan(2x). And one thing, what did you mean by "Now we know that cos(arctanx)=11+4x2\cos(\arctan x)=\frac{1}{\sqrt{1+4x^2}}"? Could you elaborate?

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

That was my typing mistake, also Letarctan(2x)=θ arctan(2x)=\theta, hence 2x=tan(θ)2x=tan(\theta), hence cos(θ)=11+4x2cos(\theta)=\frac{1}{\sqrt{1+4{x}^{2}}}

cos(arctan(2x))=11+4x2\Rightarrow cos(arctan(2x))=\frac{1}{\sqrt{1+4{x}^{2}}}

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

Solutionofproblem14Solution\quad of\quad problem\quad 14

Lemma

π2π2ln(a2cos2x+b2sin2x)dx=2πln(a+b2)\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 :

I(a)=π2π2ln(a2cos2x+b2sin2x)dx\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 aa we get :

I(a)=2aπ2π2cos2xa2cos2x+b2sin2xdx\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)=ttan(x)=t to get our integral as :

I(a)=2aπ2π2dt(a2+b2t2)(1+t2)\displaystyle {I}^{'}(a)=2a\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \frac { dt }{ ({ a }^{ 2 }+{ b }^{ 2 }{ t }^{ 2 })(1+{ t }^{ 2 }) } }

=2aπ2π2dt(a2+b2t2)(1+t2)=2ab2a2(π2π2b2dta2+b2t2π2π2dt1+t2)\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 :

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

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

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

Hence I(a)=2πln(a+b2)\displaystyle I(a)=2\pi ln(\frac{a+b}{2})

In our integral in the given question put x=1sinθ2x=\frac{1-sin\theta }{2} to get the integral as :

I=π2π2ln(7sin(θ))dxπ2π2ln(5+sin(θ))dxI= \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 abf(x)dx=abf(a+bx)dx\displaystyle \int _{ a }^{ b }{ f(x)dx } =\int _{ a }^{ b }{ f(a+b-x)dx } we get :

I=π2π2ln(7+sin(θ))dxπ2π2ln(5sin(θ))dx\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 :

I=12(π2π2ln(49sin2θ)dxπ2π2ln(25sin2(θ))dx)\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 } )

I=12(π2π2ln(49cos2θ+48sin2θ)dxπ2π2ln(25cos2θ+24sin2θ)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 :

I=12(2πln(7+432)2πln(5+262))=πln(7+435+26)\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 } } )

Problem15Problem\quad 15

Evaluate I=011ln(x)+11xdx\displaystyle I=\int _{ 0 }^{ 1 }{ \frac { 1 }{ ln(x) } +\frac { 1 }{ 1-x } dx }

Ronak Agarwal - 4 years, 7 months ago

Log in to reply

According to timeline, you're faster than @jatin yadav to post your solution, then the right to propose a new problem is yours.

Anastasiya Romanova - 4 years, 7 months ago

Log in to reply

_ Solution of problem 14 _

I=01ln(3+x3x)1x(1x2)dxI = \displaystyle \int_{0}^{1} \ln \bigg(\dfrac{3+x}{3-x} \bigg) \dfrac{1}{\sqrt{x(1-x^2)}} {\mathrm dx}

= 01ln(1+x/3)1x(1x)dx\displaystyle \int_{0}^{1} \ln(1 + x/3) \dfrac{1}{\sqrt{x(1-x)}} {\mathrm dx}

= 0π/22ln(1+13sin2θ)dθ0π/22ln(113sin2θ)dθ\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 0π/2ln(1+asin2θ)=πln(1+a+12)\displaystyle \int_{0}^{\pi/2} \ln(1+a \sin^2 \theta) = \pi \ln \bigg(\dfrac{1+\sqrt{a+1}}{2}\bigg)