Brilliant Integration Contest - Season 1 (Part 2)

This is Brilliant Integration Contest - Season 1 (Part 2) as a continuation of the previous contest (Part 1). There is a major change in the rules of contest, so please read all of them carefully before take part in this contest.

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.

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]

Remember, put them separately.

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

Okay, let the contest part 2 begin!

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

Note by Anastasiya Romanova
4 years, 11 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

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 \( ... \) or \[ ... \] 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

PROBLEM 33

Find 0tlog(t+q)(t+1)(t+q)dt, \displaystyle \int_0^{\infty } \frac{\sqrt{t} \log (t + q)}{(t+1) (t+q)} \, dt,

where q q is a parameter.

Hence show that 0tlog(2t+1)(t+1)(2t+1)dt=π(2log(1+2)2log(2)). \displaystyle \int_0^{\infty } \frac{\sqrt{t} \log (2 t+1)}{(t+1) (2 t+1)} \, dt = \pi \left(2 \log \left(1+\sqrt{2}\right)-\sqrt{2} \log (2)\right).

My own solution is kind of complicated, there is probably an easier way.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

SOLUTION OF PROBLEM 33 :

I am surprised that no one answers this question.

Okay, here is the solution. Setting t=x2t=x^2 and followed by partial fractions decomposition, we have 0tln(t+q)(t+1)(t+q)dt=20x2ln(x2+q)(x2+1)(x2+q)dx=2qq10ln(x2+q)x2+qdx+21q0ln(x2+q)x2+1dx=2qq1I+21qJ\begin{aligned} \int_0^\infty\frac{\sqrt{t}\ln(t+q)}{(t+1)(t+q)}\,dt&=2\int_0^\infty \frac{x^2\ln\left(x^2+q\right)}{\left(x^2+1\right)\left(x^2+q\right)}\,dx\\ &=\frac{2q}{q-1}\int_0^\infty \frac{\ln\left(x^2+q\right)}{x^2+q}\,dx+\frac{2}{1-q}\int_0^\infty \frac{\ln\left(x^2+q\right)}{x^2+1}\,dx\\ &=\frac{2q}{q-1}I+\frac{2}{1-q}J \end{aligned} Setting x=qtanθx=\sqrt{q}\tan\theta to II, we have I=1q0π/2lnqdθ+1q0π/2ln(1+tan2θ)dθ=πlnq2q2q0π/2ln(cosθ)dθ=πlnq2q+πln2q=πln(2q)q\begin{aligned} I&=\frac{1}{\sqrt{q}}\int_0^{\pi/2}\ln q\,\,d\theta+\frac{1}{\sqrt{q}}\int_0^{\pi/2}\ln\left(1+\tan^2\theta\right)\,d\theta\\ &=\frac{\pi\ln q}{2\sqrt{q}}-\frac{2}{\sqrt{q}}\int_0^{\pi/2}\ln\left(\cos\theta\right)\,d\theta\\ &=\frac{\pi\ln q}{2\sqrt{q}}+\frac{\pi\ln2}{\sqrt{q}}\\ &=\frac{\pi\ln \left(2\sqrt{q}\right)}{\sqrt{q}} \end{aligned} Differentiating JJ w.r.t. qq, we have J(q)=0dx(x2+q)(x2+1)=11q0dxx2+qsetx=qtanθ11q0dxx2+1=π2q(1q)π2(1q)J(q)=πdq2q(1q)setx=q+π2ln(1q)=π2ln(1+q1q)+π2ln(1q)+C=π2ln((1+q)(1q)1q)+C=πln(1+q)+C\begin{aligned} J'(q)&=\int_0^\infty \frac{dx}{\left(x^2+q\right)\left(x^2+1\right)}\\ &=\frac{1}{1-q}\underbrace{\int_0^\infty \frac{dx}{x^2+q}}_{\color{#D61F06}{\text{set}\,x=\sqrt{q}\tan\theta}}-\frac{1}{1-q}\int_0^\infty \frac{dx}{x^2+1}\\ &=\frac{\pi}{2\sqrt{q}\,(1-q)}-\frac{\pi}{2(1-q)}\\ J(q)&=\pi\underbrace{\int \frac{dq}{2\sqrt{q}\,(1-q)}}_{\color{#D61F06}{\text{set}\,x=\sqrt{q}}}+\frac{\pi}{2}\ln(1-q)\\ &=\frac{\pi}{2}\ln\left(\frac{1+\sqrt{q}}{1-\sqrt{q}}\right)+\frac{\pi}{2}\ln(1-q)+C\\ &=\frac{\pi}{2}\ln\left(\frac{\left(1+\sqrt{q}\right)(1-q)}{1-\sqrt{q}}\right)+C\\ &=\pi\ln\left(1+\sqrt{q}\right)+C \end{aligned} Now, to determine the value of CC, let us evaluate J(0)J(0). J(0)=20lnxx2+1dxx=tanθ=20π/2ln(tanθ)dθ=20π/2ln(sinθ)dθ20π/2ln(cosθ)dθ=0\begin{aligned} J(0)&=2\int_0^\infty \frac{\ln x}{x^2+1}\,dx\qquad\Rightarrow\qquad x=\tan\theta\\ &=2\int_0^{\pi/2}\ln(\tan\theta)\,d\theta\\ &=2\int_0^{\pi/2}\ln(\sin\theta)\,d\theta-2\int_0^{\pi/2}\ln(\cos\theta)\,d\theta\\ &=0 \end{aligned} Since J(0)=0J(0)=0, then C=0C=0. Hence 0tln(t+q)(t+1)(t+q)dt=2πqq1ln(2q)+2π1qln(1+q)=π1q[2ln(1+q)qln(4q)]\begin{aligned} \int_0^\infty\frac{\sqrt{t}\ln(t+q)}{(t+1)(t+q)}\,dt &=\frac{2\pi\sqrt{q}}{q-1}\ln \left(2\sqrt{q}\right)+\frac{2\pi}{1-q}\ln\left(1+\sqrt{q}\right)\\ &=\frac{\pi}{1-q}\left[2\ln\left(1+\sqrt{q}\right)-\sqrt{q}\ln \left(4q\right)\right] \end{aligned} It holds for q>0q>0.

Although I know how to evaluate the second integral, i.e. use q=12q=\frac{1}{2}, but I refuse to answer it since it violates the rules. Sorry... ¨\quad\ddot\smile

P.S. PROBLEM 34 will be posted here: Brilliant Integration Contest - Season 1 (Part 3)

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

Nice solution Anastasiya. I might need to think of a more difficult one for you next time ;).

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

What do you mean by answering it violates the rules.

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

@Ronak Agarwal According to rule no. 7, I consider this as double integrals. You may ask only one question in each thread. Although, it also doesn't matter since I am able to prove it too. Anyway, I've posted PROBLEM 34 in the new note. You may also take a look there. @Ruben Doornenbal too.

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

Hint: the value of the first integral is

π2log(q+1)qlog(4q)1q. \displaystyle \pi \frac{2 \log \left(\sqrt{q}+1\right)-\sqrt{q}\log(4q) }{1-q}.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

PROBLEM 32

Find the closed-form expression of xsin(x+sinx)ecosx1+x2dx\int_{-\infty}^{\infty} \frac{x\sin(x+\sin x)\,e^{\cos x}}{1+x^2}\,dx

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

SOLUTION 32

We have

I=0xeix+eixdx1+x2=k01k!0xei(k+1)xdx1+x2=k01k!πie(k+1)=πe1+e1. \displaystyle I = \Im \int_0^\infty \frac{x e^{ix + e^{ix}} dx}{1+x^2} = \Im \sum\limits_{k \geq 0} \frac{1}{k!} \int_0^\infty \frac{x e^{i(k+1)x} dx}{1+x^2} = \Im \sum\limits_{k \geq 0} \frac{1}{k!} \pi i e^{-(k+1)} = \pi e^{-1+e^{-1}}.

Here we made use of the fact that for positive aa ,

0xeiaxdx1+x2=1idda0eiaxdx1+x2=1iddaπea=iπea. \displaystyle \int_0^\infty \frac{x e^{i a x} dx}{1+x^2} = \frac 1 i \frac{d}{da} \int_0^\infty \frac{e^{i a x} dx}{1+x^2} = \frac 1 i \frac{d}{da} \pi e^{-a} = i \pi e^{-a}.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

I will post problem 33 when I get home.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

@Ruben Doornenbal Post PROBLEM 33 here: Brilliant Integration Contest - Season 1 (Part 3). This note is already full.

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

PROBLEM -31 Compute the integral 0πln(12acos(x)+a2)dx \int_{0}^{\pi} \ln(1-2a\cos(x)+a^2)dx , a>1 a > 1

Shivang Jindal - 4 years, 11 months ago

Log in to reply

I posted this problem, because there are numerous elegant ways to solve this problem. There exist a simple solution, by using Riemann integral[which i not commonly used], making this problem extremely beautiful. Also, this problem can be done elegantly using Physics[gravitation]. I will post both the solutions, if anyone wants.

Shivang Jindal - 4 years, 11 months ago

Log in to reply

I think I have seen this somewhere before. Is it from the book "Inside Interesting Integrals" or from the book "Irresistible integrals"

Rohan Shinde - 11 months ago

Log in to reply

@Shivang Jindal Can you post your solutions for this integral when you have time?

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Please do share them for us here :)

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

I am very interested in these solutions!

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

SOLUTION OF PROBLEM 31

I(a)=0πln(12acosx+a2)  dx={0,a<12πlna,a>1I(a)=\int_0^\pi\,\ln\left(1-2a\cos x+a^2\right)\;dx=\begin{cases} 0 &,\quad|a| < 1\\[12pt] 2\pi\ln|a| &,\quad |a| > 1 \end{cases}

For a complete proof, you may refer to my answer on Math S.E.

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

Problem30Problem\quad 30

Find 1xx0.5xdx\displaystyle \int _{ 1 }^{ \infty }{ \frac { x-\left\lfloor x \right\rfloor -0.5 }{ x } dx }

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

I=limnr=1r=n1rr+1xr12xdx I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} \int_{r}^{r+1} \frac{x-r-\frac{1}{2}}{x} dx I=limnr=1r=n11(r+12)ln(r+1r) I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 1-(r+\frac{1}{2})\ln(\frac{r+1}{r}) 2I=limnr=1r=n12(2r+1)ln(r+1r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+1)\ln(\frac{r+1}{r}) 2I=limnr=1r=n12(2r+2)ln(r+1r)+ln(r+1r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(\frac{r+1}{r}) + \ln(\frac{r+1}{r}) 2I=limnr=1r=n12(2r+2)ln(r+1)+(2r+2)ln(r)+ln(r+1r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(r+1) +(2r+2)\ln(r)+ \ln(\frac{r+1}{r}) 2I=limnr=1r=n122((r+1)ln(r+1)rln(r))+ln(r+1)+ln(r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r) 2I=limnr=1r=n122((r+1)ln(r+1)rln(r))+ln(r+1)+ln(r) 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r) 2I=limn2(n1)2nln(n)+2ln(n!)ln(n) 2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+2\ln(n!)-\ln(n) 2I=limn2(n1)2nln(n)+(2n+1)ln(n)2n+ln(2π)ln(n) 2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+(2n+1)\ln(n)-2n+\ln(2\pi)-\ln(n) 2I=ln(2π)2 2I = \ln(2\pi)-2 I=ln(2π)21 I =\frac{\ln(2\pi)}{2}-1

Shivang Jindal - 4 years, 11 months ago

Log in to reply

Trick , is to break the integral from (1,2),(2,3)...(n1,n) (1,2),(2,3)...(n-1,n) . Then, we compute the sum in terms of n n . and then use Stirling approximation :) .

Shivang Jindal - 4 years, 11 months ago

Log in to reply

Exactly, you got it perfectly right.

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

This one is easy too :) ln2π21\frac{\ln2\pi}{2}-1

@Shivang Jindal : Sorry, I was kidding & I am busy right now so I have no time to write down my answer. Could you elaborate yours then you're good to go (propose your problem). Sorry for the inconvenience...

Tunk-Fey Ariawan - 4 years, 11 months ago

Log in to reply

Expecting a question from you @Tunk-Fey Ariawan , also please give a proof of your answer.

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

PROBLEM 29

Oops, I just noticed that the one I posted contains a trigamma function (not allowed by the rules). Here's a better one:

01x2tan1(x)2dx=G3+148(4π)2π12ln2. \displaystyle \int_0^1 x^2 \tan ^{-1}(x)^2 \, dx = \frac G 3 + \frac {1}{ 48} (4 - \pi)^2 - \frac {\pi}{12} \ln 2.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Solutiontoproblem29Solution\quad to\quad problem\quad 29

This one's easy.

Apply integration by parts u=atan(x)2u={atan(x)}^{2} and dv=x2dxdv={x}^{2}dx

I=x3tan1(x)23012301x3tan1x1+x2dx\displaystyle I=\frac { { x }^{ 3 }{ { tan }^{ -1 }(x) }^{ 2 } }{ 3 } \overset { 1 }{ \underset { 0 }{ | } } -\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \frac { { x }^{ 3 }{ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } dx }

I=π2482301(x3+xx)tan1x1+x2dx\displaystyle I=\frac { { \pi }^{ 2 } }{ 48 } -\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \frac { ({ x }^{ 3 }+x-x){ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } dx }

I=π2482301xtan1xdx+2301xtan1x1+x2dx\displaystyle I=\frac { { \pi }^{ 2 } }{ 48 } -\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ x{ tan }^{ -1 }xdx } +\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \frac { x{ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } dx }

J=01xtan1xdx\displaystyle J=\int _{ 0 }^{ 1 }{ x{ tan }^{ -1 }xdx }

Integration by parts u=tan1(x),dv=xdxu=tan^{-1}(x),dv=xdx

J=x2tan1x2011201x2dx1+x2\displaystyle J=\frac { { x }^{ 2 }{ tan }^{ -1 }x }{ 2 } \overset { 1 }{ \underset { 0 }{ | } } -\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 }dx }{ 1+{ x }^{ 2 } } }

Rest is trivial and it evaluates out to be :

J=π412J=\frac{\pi}{4}-\frac{1}{2}

K=01xtan1xdx1+x2\displaystyle K=\int _{ 0 }^{ 1 }{ \frac { x{ tan }^{ -1 }xdx }{ 1+{ x }^{ 2 } } }

Put x=tan(θ)x=tan(\theta) to get the integral as :

K=0π4θtanθdθ\displaystyle K=\int _{ 0 }^{ \frac{\pi}{4} }{ \theta tan\theta d\theta }

Integration by parts dv=tanθdθ,u=θdv=tan\theta d\theta , u=\theta

K=θln(secθ)0π4+0π4ln(cosθ)dθ\displaystyle K=\theta ln(sec\theta )\overset { \frac { \pi }{ 4 } }{ \underset { 0 }{ | } } +\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cos\theta )d\theta }

K=πln(2)8+0π4ln(cosθ)dθ\displaystyle K=\frac { \pi ln(2) }{ 8 } +\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cos\theta )d\theta }

Now M=0π4ln(cosθ)dθ=π4π2ln(sinθ)dθ=0π2ln(sinθ)dθ0π4ln(sinθ)dθ\displaystyle M=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cos\theta )d\theta } =\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta } -\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(sin\theta )d\theta }

Adding both these forms we get :

M=120π4ln(cotθ)dθ+0π2ln(sinθ)dθ\displaystyle M=\frac{1}{2}\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cot\theta )d\theta } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta }

M=G2πln(2)4M=\frac { G }{ 2 } -\frac { \pi ln(2) }{ 4 }

Here I have two general results :

πln(2)2=0π2ln(sinθ)dθ,0π4ln(cotθ)dθ=G\displaystyle \frac { -\pi ln(2) }{ 2 } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta },\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cot\theta )d\theta } =G

Where GG is the catalan's constant,.

Finally we get K=G2πln(2)8K= \frac{G}{2}-\frac{\pi ln(2)}{8}

Using all these values we finally get II as :

I=π248+G3+13π6πln212\large I=\frac{{\pi}^{2}}{48}+\frac{G}{3}+\frac{1}{3}-\frac{\pi}{6}-\frac{\pi ln2}{12}

I=(4π)248+G3πln(2)12\Rightarrow \large \boxed{I=\frac { { (4-\pi ) }^{ 2 } }{ 48 } +\frac { G }{ 3 } -\frac { \pi ln(2) }{ 12 } }

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

Problem 28

Prove

sinh2xcos2xsinhπx dx=sin2cos2+cosh2\begin{aligned} \int_{-\infty}^{\infty} \frac{\sinh 2x\cos 2x}{\sinh \pi x} \ dx = \frac{\sin 2}{\cos 2 + \cosh 2} \end{aligned}

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

Solution 28

The integral equals

I=dx(e2xe2x)eπxe2ixe2πx1.\displaystyle I = \Re \int_{-\infty}^\infty dx \frac{\left(e^{2x} - e^{- 2x} \right) e^{\pi x} e^{2 i x}}{e^{2 \pi x} - 1}.

Substitute eπx=u e^{\pi x} = u . We get

I=J+J, \displaystyle I = J_+ - J_-,

where

J±=1π0dxu2(i±1)/πu21=12cot[π2(2(i±1)/π+1)]=12tan(i±1).\displaystyle \begin{aligned} J_\pm &= \Re \frac 1 \pi \int_0^\infty dx \frac{u^{2( i\pm1)/\pi}}{u^2 - 1} \\&= -\Re \frac 1 2 \cot\left[\frac \pi 2 \left(2(i\pm1)/\pi + 1\right) \right] \\&= \Re\frac 1 2 \tan(i \pm 1). \end{aligned}

Here we made use of the well-known integral

PV0xa11xb=πbcotπab.\displaystyle PV \int_0^\infty \frac{x^{a-1}}{1-x^b} = \frac \pi b \cot \frac{\pi a}{b}.

Now using the identity

tanA+B2=sinA+sinBcosA+cosB\displaystyle \tan\frac{A+B}{2} = \frac{\sin A + \sin B}{\cos A + \cos B}

gives

J±=±12sin2cosh2+cos2,\displaystyle J_\pm = \pm \frac 1 2 \frac{\sin 2}{\cosh 2 + \cos 2},

which gives the desired result.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Wait!? For PV0xa11xbdx=πbcot(πab)PV\int_0^\infty \frac{x^{a-1}}{1-x^b}\,dx=\frac{\pi}{b}\cot\left(\frac{\pi a}{b}\right) could you prove it without using contour integration or residue method? See the rules.

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

@Anastasiya Romanova Of course, my dear. It is clear that we can take b=1 b = 1 in the proof. The general result follows from a substitution. Separate the integrals over (0,1) (0,1) and over (1,) (1,\infty) . Put u=1/x u = 1/x in the second integral. The result is

01dxxa1xa1x=ψ(1a)ψ(a)=πcotπa, \displaystyle \int_0^1 dx \frac{x^{a-1} - x^{-a}}{1-x} = \psi(1-a) - \psi(a) = \pi \cot \pi a,

as was to be proven. Here we used a result derived by real methods here.

You have sharp eye for integrals that I normally derive with residues :p

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

@Ruben Doornenbal OK, I accept your explanation. You may propose your problem but I have to sleep now. It's already late night here. Maybe someone else will answer your problem. So, in the mean time, you fight with them. Good night! 👋(>‿◠)

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

PROBLEM 27

My last two integrals were clearly too easy. By finding an antiderivative or otherwise, show that

0dxln2tanhx=74ζ(3). \int_0^{\infty} dx\, \ln^2 \tanh x = \frac{7}{4}\zeta{(3)}.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Solution of Problem 27

Set t=tanhxt=\tanh x, we have

0ln2(tanhx)dx=01ln2t1t2dt=01n=0t2nln2tdt=n=001t2nln2tdt=2n=01(2n+1)3see solution of Problem 13=2[n=11n3n=11(2n)3]=74n=11n3=74ζ(3)\begin{aligned} \int_0^\infty\ln^2(\tanh x)\,dx&=\int_0^1\frac{\ln^2t}{1-t^2}\,dt\\ &=\int_0^1\sum_{n=0}^\infty t^{2n}\ln^2t\,dt\\ &=\sum_{n=0}^\infty\int_0^1 t^{2n}\ln^2t\,dt\\ &=2\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\qquad\Rightarrow\qquad\text{see solution of Problem 13}\\ &=2\left[\sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{1}{(2n)^3}\right]\\ &=\frac{7}{4}\sum_{n=1}^\infty\frac{1}{n^3}\\ &=\frac{7}{4}\zeta(3) \end{aligned}

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

PROBLEM 26 :

Prove

0lnxcoshxdx=π2ln(Γ4(34)π)\int_0^\infty\frac{\ln x}{\cosh x}\,dx=\frac{\pi}{2}\ln\left(\frac{\Gamma^4\left(\frac{3}{4}\right)}{\pi}\right)

P.S. You may use any well-known expressions.

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

SOLUTION 26

Consider

I(a)=0dxxa1coshx=2k0(1)k0dxxa1e(2k+1)x=2Γ(a)β(a). \displaystyle I(a) = \int_0^\infty dx \frac{x^{a-1}}{\cosh x} = 2 \sum \limits_{k \geq 0} (-1)^k \int_0^\infty dx x^{a-1} e^{-(2k+1)x} = 2 \Gamma(a) \beta(a).

Our integral is

I(1)=2Γ(1)β(1)+2Γ(1)β(1)=2(γ)(π/4)+2π4[γ+2ln2+3lnπ4lnΓ14]. \displaystyle I'(1) = 2 \Gamma'(1)\beta(1) + 2 \Gamma(1) \beta'(1) = 2(-\gamma)(\pi/4) + 2 \frac \pi 4 \left[\gamma + 2 \ln 2 + 3 \ln \pi - 4 \ln \Gamma \frac 1 4 \right].

Here we used a result from Mathworld. Using the Euler reflection formula,

Γ(1/4)=π2(Γ(3/4))1. \displaystyle \Gamma(1/4) = \pi \sqrt 2 (\Gamma(3/4))^{-1}.

Collecting all the terms gives I(1)=π2lnπ+2πlnΓ(3/4), \displaystyle I'(1) = -\frac \pi 2 \ln \pi + 2\pi \ln \Gamma(3/4),

which equals the stated result.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Problem25Problem\quad 25

Find 01ln(ln(1x))dx\displaystyle \int _{ 0 }^{ 1 }{ ln(ln(\frac { 1 }{ x } ))dx }

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

A well-known problem. Set t=ln(1x)t=\ln\left(\frac{1}{x}\right), we have

01lnln(1x)dx=0etlntdt\int_0^1 \ln \ln \left(\frac{1}{x}\right)\,dx=\int_0^\infty e^{-t}\ln t\, dt

The last expression is derivative of gamma function where its parameter a=1a=1, we have

lima1aΓ(a)=lima1a0ta1etdt=lima1ψ(a)Γ(a)=γ\lim_{a\to1}\partial_a\,\Gamma(a)=\lim_{a\to1}\partial_a\,\int_0^\infty t^{a-1}e^{-t}\, dt=\lim_{a\to1}\psi(a)\Gamma(a)=-\gamma

where ψ(x)\psi(x) is the digamma function and γ\gamma is Euler–Mascheroni constant.

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

How can you prove that derivative of gamma function at one is equal to euler's mascheroni constant. @Anastasiya Romanova

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

@Ronak Agarwal from definition of the digamma function we have

ψ(x)=ddxlnΓ(x)=Γ(x)Γ(x)\psi(x)=\frac{d}{dx}\ln\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}

Hence

Γ(x)=Γ(x)ψ(x)\Gamma'(x)=\Gamma(x)\psi(x)

and

Γ(1)=ψ(1)\Gamma'(1)=\psi(1)

Now we may use any identities related for the digamma function, for example

ψ(n)=Hn1γ\psi(n)=H_{n-1}-\gamma

where Hn1H_{n-1} is the harmonic number and by definition H0=0H_0=0 or you may refer to this and post on Math S.E. where I also contribute an answer there.

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

PROBLEM 24

01arcsechxarcsinxdx=π28ln2. \displaystyle \int_0^1 \operatorname{arcsech} x \operatorname{arcsin} xdx =\frac{\pi^2}{8} - \ln 2.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Solutionofproblem24Solution\quad of\quad problem\quad 24

First note that :

arcsech(x)=ln(1+1x2x)arcsech(x)=ln(\frac{1+\sqrt{1-{x}^{2}}}{x})

In our integral put x=sin(θ)x=sin(\theta)

I=0π2θcosθln(1+cosθsinθ)dθ\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \theta cos\theta ln(\frac { 1+cos\theta }{ sin\theta } ) } d\theta

Applying integration by parts we get , u=ln(1+cosθsinθ),dv=θcosθdθu=ln(\frac{1+cos\theta }{sin\theta}),dv=\theta cos\theta d\theta

I=(θsinθ+cosθ)ln(1+cosθsinθ)0π2+0π2(θsinθ+cosθ)cosecθdθ\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\theta sin\theta +cos\theta )cosec\theta d\theta }

I=(θsinθ+cosθ)ln(1+cosθsinθ)+θ22+ln(sinθ)0π2\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )+\frac { { \theta }^{ 2 } }{ 2 } +ln(sin\theta )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } }

Which on evaluating we get :

I=π28ln(2)I=\frac { { \pi }^{ 2 } }{ 8 } -ln(2)

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

Can you elaborate the first line please , it will be great to learn from you

U Z - 4 years, 10 months ago

Log in to reply

Problem23Problem\quad 23

Find 0π4ln(tan(x))dx\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(tan(x))dx }

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

@Ronak Agarwal @Ruben Doornenbal The answer given is wrong!!! It should be -G!!! You must have forgotten the negative sign.......

Aaghaz Mahajan - 1 year, 1 month ago

Log in to reply

Solution 23

A well-known problem. Sub tanx=u \tan x = u to get

01dudulnu1+u2=k0(1)k01dulnuu2k=k0(1)k1(2k+1)2=G. \displaystyle \int_0^1 du \frac{du \ln u}{1+u^2} = \sum_{k \geq 0} (-1)^k \int_0^1 du \ln u \, u^{2k} = \sum_{k \geq 0} (-1)^k \frac{1}{(2k+1)^2} = G.

The penultimate equality follows from integration by parts.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Sir can you elaborate I did'nt understood this one @Ruben Doornenbal

U Z - 4 years, 10 months ago

Log in to reply

@U Z The idea is to expand the factor 11+u2 \displaystyle \frac{1}{1+u^2} in a geometric series and interchange summation and integration. The last equality is just the definition of Catalan's constant.

Ruben Doornenbal - 4 years, 10 months ago

Log in to reply

@Ruben Doornenbal Can we have problem 24?

Samuel Jones - 4 years, 11 months ago

Log in to reply

PROBLEM 22

This one is particularly beautiful, in my opinion.

0axdxcosxcos(ax)=asinalnseca. \displaystyle \int_0^a \frac{x dx}{\cos x \cos(a-x)} = \frac{a}{\sin a} \ln \sec a.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

SolutionofProblem22Solution\quad of\quad Problem\quad 22

I=0axdxcos(x)cos(ax)=0a(ax)dxcos(x)cos(ax)I=\displaystyle \int _{ 0 }^{ a }{ \frac { xdx }{ cos(x)cos(a-x) } } =\int _{ 0 }^{ a }{ \frac { (a-x)dx }{ cos(x)cos(a-x) } }

Adding these two forms we get :

I=a20adxcos(x)cos(ax)I=\displaystyle \frac { a }{ 2 } \int _{ 0 }^{ a }{ \frac { dx }{ cos(x)cos(a-x) } }

Multiplying and dividing by sin(a)sin(a) we get :

I=a2sin(a)0asin(x+(ax))dxcos(x)cos(ax)I=\displaystyle \frac { a }{ 2sin(a) } \int _{ 0 }^{ a }{ \frac { sin(x+(a-x))dx }{ cos(x)cos(a-x) } }

I=a2sin(a)0a(tan(x)+tan(ax))dxI=\displaystyle \frac { a }{ 2sin(a) } \int _{ 0 }^{ a }{ (tan(x)+tan(a-x))dx }

Also since 0atan(x)dx=0atan(ax)dx\displaystyle \int _{ 0 }^{ a }{ tan(x)dx } =\int _{ 0 }^{ a }{ tan(a-x)dx }

We get I=asin(a)0atan(x)dxI=\displaystyle \frac{a}{sin(a)}\int _{ 0 }^{ a }{ tan(x)dx }

I=asin(a)ln(sec(a))I=\frac { a }{ sin(a) } ln(sec(a))

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

PROBLEM 21 :

Show that 0dxx4+2cos(2θ)x2+1=π4cosθ\int_0^\infty \frac{dx}{x^4+2\cos(2\theta)\,x^2+1}=\frac{\pi}{4\cos\theta}

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

We have

0dxx4+ax2+b2=π2b2b+a, \displaystyle \int_0^ {\infty} \frac{dx}{x^4 + a x^2 + b^2} = \frac{\pi}{2b \sqrt{2b+a}},

which I proved on MSE. Plugging in a=2cos2θ a = 2 \cos2 \theta and using that 2cos2(θ)=1+cos2θ2 \cos^2(\theta) = 1 + \cos 2\theta immediately gives the answer.

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Expecting a question from you @Ruben Doornenbal

Ronak Agarwal - 4 years, 11 months ago

Log in to reply

Oh I know you're now. You're user111187. I thought you're an old man. Haha

Nice to meet you here Ruben. It seems you'll be a tough opponent because you're a Math SE and I&S user. ¨\ddot\smile

Anastasiya Romanova - 4 years, 11 months ago

Log in to reply

@Anastasiya Romanova Yep, this will be good :)

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Problem 20

01sinh1(x)log[(21)x+1]xdx=log(2)log(1+2)π224 \displaystyle \int_0^1 \frac{\sinh ^{-1}(x)-\log \left[\left(\sqrt{2}-1\right) \sqrt{x}+1\right]}{x} \, dx = \log (2) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{24}

Ruben Doornenbal - 4 years, 11 months ago

Log in to reply

Solving this problem is really tedious job. Honestly, I'm unwilling to answer it too (even if I know how to solve it). I don't know what is the OP's motivation by posting two well-defined integrals in a single problem. If the OP has an elementary and a clever method than mine, please do share to us. Okay, here is an attempt using a cannon.

Let split the integral into two parts

IJ=01arcsinhxxdx01ln(1+(21)x)xdxI-J=\int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx-\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx

Perform integration by parts for II by taking u=arcsinhxu={\rm{arcsinh}\,} x and dv=dxxdv=\dfrac{dx}{x}.

01arcsinhxxdx=arcsinhxlnx0101lnx1+x2dxx=tantI=0π/4ln(cost)ln(sint)costdt=0π/4ln(cost)costdt0π/4ln(sint)costdt=120π/4ln(1sin2t)costdt0π/4ln(sint)costdt\begin{aligned} \int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx&={\rm{arcsinh}\,} x\,\ln x\bigg|_0^1-\int_0^1\frac{\ln x}{\sqrt{1+x^2}}\,dx\qquad\Rightarrow\qquad x=\tan t\\ I&=\int_0^{\pi/4}\frac{\ln(\cos t)-\ln (\sin t)}{\cos t}\,dt\\ &=\int_0^{\pi/4}\frac{\ln(\cos t)}{\cos t}\,dt-\int_0^{\pi/4}\frac{\ln(\sin t)}{\cos t}\,dt\\ &=\frac{1}{2}\int_0^{\pi/4}\frac{\ln(1-\sin^2 t)}{\cos t}\,dt-\int_0^{\pi/4}\frac{\ln(\sin t)}{\cos t}\,dt\\ \end{aligned}

Putting y=sinty=\sin t and a=12a=\dfrac{1}{\sqrt{2}}, we get

01arcsinhxxdx=0aln(1y2)1y2dy0alny1y2dy=0aln(1y)+ln(1+y)(1y)(1+y)dy0alny(1y)(1+y)dy=I1+I2\begin{aligned} \int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx&=\int_0^{a}\frac{\ln (1-y^2)}{1-y^2}\,dy-\int_0^{a}\frac{\ln y}{1-y^2}\,dy\\ &=\int_0^{a}\frac{\ln (1-y)+\ln(1+y)}{(1-y)(1+y)}\,dy-\int_0^{a}\frac{\ln y}{(1-y)(1+y)}\,dy\\ &=I_1+I_2 \end{aligned}

Performing partial fractions decomposition we get

I1=120aln(1+y)1+ydy+120aln(1y)1+ydy+120aln(1+y)1ydy+120aln(1y)1ydy=ln2(1+a)4+120aln(1y)1+ydy+120aln(1+y)1ydyln2(1a)4=14ln2(1+a1a)+120aln(1y)1+ydy+120aln(1+y)1ydy2z=1+y=ln2(1+2)+12cbln(22z)zdz+12cblnz1zdz=ln2(1+2)+ln22cbdzz+12cbln(1z)zdz+121c1bln(1z)zdz=ln2(1+2)+ln22ln(1+22)+12cbln(1z)zdz+121c1bln(1z)zdz\begin{aligned} I_1&=\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1-y}\,dy\\ &=\frac{\ln^2 (1+a)}{4}+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy-\frac{\ln^2 (1-a)}{4}\\ &=\frac{1}{4}\ln^2\left(\frac{1+a}{1-a}\right)+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy\qquad\Rightarrow\qquad 2z=1+y\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{1}{2}\int_c^{b}\frac{\ln (2-2z)}{z}\,dz+\frac{1}{2}\int_c^{b}\frac{\ln z}{1-z}\,dz\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{\ln2}{2}\int_c^{b}\frac{dz}{z}+\frac{1}{2}\int_c^{b}\frac{\ln (1-z)}{z}\,dz+\frac{1}{2}\int_{1-c}^{1-b}\frac{\ln (1-z)}{z}\,dz\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{\ln2}{2}\ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)+\frac{1}{2}\int_c^{b}\frac{\ln (1-z)}{z}\,dz+\frac{1}{2}\int_{1-c}^{1-b}\frac{\ln (1-z)}{z}\,dz\\ \end{aligned}