# 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) :

PROBLEM xxx (number of problem) :

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
6 years, 4 months ago

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Thanks for doing this. There is a lot to learn from these integration questions that you have shared.

Staff - 6 years, 4 months ago

Thank you for your help. You're too kind to me. I really appreciate it $\quad$ $\ddot\smile$

- 6 years, 4 months ago

Problem 20

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

- 6 years, 4 months ago

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

$I-J=\int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx-\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx$

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

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

\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

\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}

and

\begin{aligned} I_2&=\frac{1}{2}\int_0^{a}\frac{\ln y}{1-y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln y}{1+y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy+\frac{\ln y\ln(1+y)}{2}\bigg|_0^{a}-\int_0^{a}\frac{\ln (1+y)}{y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy+\frac{\ln a\ln(1+a)}{2}-\int_0^{a}\frac{\ln (1+y)}{y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy-\int_0^{a}\frac{\ln (1+y)}{y}\,dy-\frac{\ln2}{4}\ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)\\ \end{aligned}

where $b=\dfrac{1+\sqrt{2}}{2\sqrt{2}}$ and $c=\dfrac{1}{2}$.

Now, let us evaluate $J$. Set $x=t^2$, we get

$J=\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx=2\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)t\right)}{t}\,dt$

Here is the cannon, recall a special function dilogarithm.

$\operatorname{Li}_2(z)=-\int_0^z\frac{\ln(1 - t)}{t}\,dt$

Hence, the rest integrals can be easily evaluated by using dilogarithm and an elementary substitution, i.e. $t=kx$, where $k$ is a constant. We may also utilize these identities \begin{aligned} \operatorname{Li}_2(z)+\operatorname{Li}_2(-z)&=\frac{1}{2}\operatorname{Li}_2(z^2)\\ \operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1-\frac{1}{z}\right)&=-\frac{\ln^2z}{2}\\ \operatorname{Li}_2(z)+\operatorname{Li}_2(1-z)&=\frac{{\pi}^2}{6}-\ln z \cdot\ln(1-z) \end{aligned} and special values \begin{aligned} \operatorname{Li}_2(-1)&=-\frac{{\pi}^2}{12}\\ \operatorname{Li}_2(0)&=0\\ \operatorname{Li}_2\left(\frac{1}{2}\right)&=\frac{{\pi}^2}{12}-\frac{\ln^2 2}{2}\\ \operatorname{Li}_2(1)&=\frac{{\pi}^2}{6} \end{aligned} Performing a cumbersome and a tedious calculation, we will get the announced result.

I hope you understand my feelings while trying to solve and to write it down. So, please do not ever post a problem like this again. LOL

- 6 years, 4 months ago

The reason for taking the difference of the two integrals is that the result is much simpler than the two integrals separately (which involve dilogarithms), which I found beautiful. It only works if both terms are exactly as they are (including the weird factor $(\sqrt 2 -1)$ ).

Your solution can be significantly simplified. Both terms in the integral have a relatively simple antiderivative in terms of dilogarithms. After plugging in the limits, it then boils down to showing

$\displaystyle \operatorname{Li}_2 (\sqrt{2} - 1) - \operatorname{Li}_2 (1 - \sqrt 2) = \frac{\pi^ 2}{8} - \frac 1 2 \log^ 2(\sqrt 2 - 1)$

using the dilogarithm identities you posted.

To find the antiderivative of the second term, just substitute $(\sqrt 2 -1) \sqrt x = u$. For the first term, substitute $u = \left( x + \sqrt{1+x^2} \right)^2$. This will reduce the integral to

$\displaystyle \int \frac{\ln u \,du}{u^2 - 1},$

which I am sure you can calculate in a few lines.

- 6 years, 4 months ago

I'm sorry, I'm a bit dizzy right now so I can't follow your comment. Could you elaborate? If I may ask, could you post your solution of this problem? Thanks.

Edit : Aha! I get it. Use this relation: ${\rm{arcsinh}\,}x=\ln\left(x+\sqrt{x^2+1}\right)$. Very clever!

- 6 years, 4 months ago

Here's a summary of what I said, maybe it helps:

The term with $\operatorname{arcsinh} x / x$ can be evaluated by substituting $u = \left( x + \sqrt{1+x^2} \right)^2$. The result is something with a dilogarithm. The other term also gives a dilogarithm, but the dilogarithm terms cancel precisely, leading to an elementary result. I hope you can appreciate the beauty of the problem :)

- 6 years, 4 months ago

PROBLEM 16 :

Prove

$\large\int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx=\frac{\pi}{2^v v\ \operatorname{B}\left(\frac{v+a+1}{2},\frac{v-a+1}{2}\right)}$

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

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

- 6 years, 4 months ago

Isn't this problem too difficult to high school students Anna? I decide to answer Problem 16 because I'm afraid if this continues till a week, this contest will lose its interest. IMHO, you should propose an easy problem so that this contest will be fun as the stated aims of it. So, here is a solution:

SOLUTION OF PROBLEM 16 :

Rewrite the integral as follows \begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac12\int_{-\large\frac\pi2}^{\large\frac\pi2}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{v-1}\cos ax\ dx\\ &=\frac1{2^{v}}\int_{-\large\frac\pi2}^{\large\frac\pi2}\left(1+e^{2ix}\right)^{v-1}e^{-i(v-1)x}\cos ax\ dx\\ &=\frac1{2^{v}}\int_{-\large\frac\pi2}^{\large\frac\pi2}\sum_{n=0}^{v-1}\binom{v-1}{n} e^{2inx}\cdot e^{-i(v-1)x}\cos ax\ dx\\ &=\frac1{2^{v}}\sum_{n=0}^{v-1}\binom{v-1}{n} \int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i(2n-v+1)x}\cos ax\ dx.\qquad\qquad\qquad\tag1 \end{aligned} Consider $f(x)=\left\{ \begin{array}{l l} e^{i\omega x} & \quad \text{for}\ -\frac\pi2 The Fourier transform of $f(x)$ is \begin{aligned} \mathscr{F}\left[f(x)\right]&=\int_{-\infty}^\infty f(x)\ e^{-i\alpha x}\ dx\\ \int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i\omega x}\cos \alpha x\ dx-i\int_{-\large\frac\pi2}^{\large\frac\pi2}e^{i\omega x}\sin \alpha x\ dx&=\int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i\omega x}\ e^{-i\alpha x}\ dx\\ &=\int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i(\omega-\alpha) x}\ dx\\ &=\left[\frac{e^{i(\omega-\alpha) x}}{i(\omega-\alpha)}\right]_{x=-\large\frac\pi2}^{\large\frac\pi2}\\ \Re\bigg(\mathscr{F}\left[f(x)\right]\bigg)&=\left[\frac{\sin(\omega-\alpha) x}{\omega-\alpha}\right]_{x=-\large\frac\pi2}^{\large\frac\pi2}\\ \int_{\large-\frac\pi2}^{\large\frac\pi2}\cos \alpha x\ dx&=\frac{2\sin(\omega-\alpha) \frac\pi2}{\omega-\alpha}.\qquad\qquad\qquad\tag2 \end{aligned} Using $(2)$, then $(1)$ turns out to be \begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac1{2^{v-1}}\sum_{n=0}^{v-1}\binom{v-1}{n} \frac{\sin(2n-v+1-a) \frac\pi2}{2n-v+1-a}\\ &=\frac1{2^{v}}\sum_{n=0}^{v-1}\binom{v-1}{n} \frac{\sin\left(n-\frac{v-1+a}2\right) \pi}{n-\frac{v-1+a}2}.\qquad\qquad\qquad\tag3\\ \end{aligned} Now, let us express $\dbinom{y}{z}$ in term of beta function that can be related to $(3)$. \begin{aligned} \binom{y}{z}&=\frac{y!}{z!(y-z)!}\\ &=\frac{y!}{\Gamma(1+z)\Gamma(1+y-z)}\\ &=\frac{y!}{z\Gamma(z)\Gamma(1-z)(y-z)\cdots(1-z)}\\ &=\frac{\sin(\pi z)}{\pi z}\cdot\frac{y!}{(y-z)\cdots(1-z)}\\ &=\frac{\sin(\pi z)}{\pi z}\sum_{n=0}^{y}\binom{y}{n}(-1)^n\frac{n}{z-n}\\ &=\sum_{n=0}^{y}\binom{y}{n}\frac{\sin\pi(z-n)}{\pi(z-n)}.\qquad\qquad\qquad\tag4 \end{aligned} Using $(4)$, then $(3)$ turns out to be \begin{aligned} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac{\pi}{2^{v}}\binom{v-1}{\frac{v-1+a}2}\\ &=\frac{\pi}{2^{v}}\frac{\Gamma(v)}{\Gamma\left(\frac{v+a+1}2\right)\Gamma\left(\frac{v-a+1}2\right)}\\ &=\frac{\pi}{2^{v}\ v\ \operatorname{B}\left(\frac{v+a+1}2,\frac{v-a+1}2\right)}\qquad\qquad\qquad\blacksquare \end{aligned}

- 6 years, 4 months ago

OK, fine. I'll post high school integral problems from now. -_-"

- 6 years, 4 months ago

No Just keep posting those hard integrals, it's challenging but we learn a lot from it

- 6 years, 4 months ago

Your solution is valid only if $v$ is an integer, whereas the identity holds in general also.

- 4 years, 4 months ago

PROBLEM 18

Show that

$\displaystyle \int_0^{\pi/4} \tan^{1/3} x dx = \frac{1}{6} \left( \pi \sqrt{3} -3\log 2\right)$

My bad, it should be $3 \log 2$ indeed. Kinshuk's result is correct. Sorry for the confusion.

- 6 years, 4 months ago

There should be 3log(2) instead of 2log(2)

- 6 years, 4 months ago

$Solution\quad of\quad problem\quad 18:\\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { tan }^{ \frac { 1 }{ 3 } }x.dx } \\ put\quad { tan }^{ \frac { 1 }{ 3 } }x=t\\ dx\quad =\quad \frac { { 3t }^{ 2 } }{ 1+{ t }^{ 6 } } .dt\\ I\quad =\quad \int { \frac { { 3t }^{ 3 } }{ 1+{ t }^{ 6 } } .dt } \\ put\quad { t }^{ 2 }=u\\ 2t.dt=du\\ I\quad =\quad \frac { 3 }{ 2 } \int { \frac { u.du }{ 1+{ u }^{ 3 } } } \\ using\quad partial\quad fraction\quad ,\quad our\quad integration\\ turns\quad out\quad to\quad be\quad :\\ I\quad =\quad \frac { -1 }{ 2 } \int { \frac { du }{ 1+u } } \quad +\quad \frac { 1 }{ 2 } \int { \frac { (1+u)du }{ { u }^{ 2 }-u+1 } } \\ after\quad solving\quad and\quad applying\quad limits:\\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { tan }^{ \frac { 1 }{ 3 } }x.dx } \quad =\quad \frac { 1 }{ 6 } (\pi \sqrt { 3 } -3\log { 2 } )$

- 6 years, 4 months ago

PROBLEM 21 :

Show that $\int_0^\infty \frac{dx}{x^4+2\cos(2\theta)\,x^2+1}=\frac{\pi}{4\cos\theta}$

- 6 years, 4 months ago

We have

$\displaystyle \int_0^ {\infty} \frac{dx}{x^4 + a x^2 + b^2} = \frac{\pi}{2b \sqrt{2b+a}},$

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

- 6 years, 4 months ago

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. $\ddot\smile$

- 6 years, 4 months ago

Yep, this will be good :)

- 6 years, 4 months ago

Expecting a question from you @Ruben Doornenbal

- 6 years, 4 months ago

$Problem\quad 23$

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

- 6 years, 4 months ago

Solution 23

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

$\displaystyle \int_0^1 du \frac{du \ln u}{1+u^2} = \sum_{k \geq 0} (-1)^k \int_0^1 du \ln u \, u^{2k} = \sum_{k \geq 0} (-1)^k \frac{1}{(2k+1)^2} = G.$

The penultimate equality follows from integration by parts.

- 6 years, 4 months ago

@Ruben Doornenbal Can we have problem 24?

- 6 years, 4 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 2 years, 6 months ago

$Problem\quad 30$

Find $\displaystyle \int _{ 1 }^{ \infty }{ \frac { x-\left\lfloor x \right\rfloor -0.5 }{ x } dx }$

- 6 years, 4 months ago

$I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} \int_{r}^{r+1} \frac{x-r-\frac{1}{2}}{x} dx$ $I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 1-(r+\frac{1}{2})\ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+1)\ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(\frac{r+1}{r}) + \ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(r+1) +(2r+2)\ln(r)+ \ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r)$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r)$ $2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+2\ln(n!)-\ln(n)$ $2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+(2n+1)\ln(n)-2n+\ln(2\pi)-\ln(n)$ $2I = \ln(2\pi)-2$ $I =\frac{\ln(2\pi)}{2}-1$

- 6 years, 4 months ago

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

@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...

- 6 years, 4 months ago

- 6 years, 4 months ago

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

- 6 years, 4 months ago

Exactly, you got it perfectly right.

- 6 years, 4 months ago

PROBLEM 22

This one is particularly beautiful, in my opinion.

$\displaystyle \int_0^a \frac{x dx}{\cos x \cos(a-x)} = \frac{a}{\sin a} \ln \sec a.$

- 6 years, 4 months ago

$Solution\quad of\quad Problem\quad 22$

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

Adding these two forms we get :

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

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

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

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

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

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

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

- 6 years, 4 months ago

PROBLEM 24

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

- 6 years, 4 months ago

$Solution\quad of\quad problem\quad 24$

First note that :

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

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

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

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

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

$\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )+\frac { { \theta }^{ 2 } }{ 2 } +ln(sin\theta )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } }$

Which on evaluating we get :

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

- 6 years, 4 months ago

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

- 6 years, 3 months ago

PROBLEM 26 :

Prove

$\int_0^\infty\frac{\ln x}{\cosh x}\,dx=\frac{\pi}{2}\ln\left(\frac{\Gamma^4\left(\frac{3}{4}\right)}{\pi}\right)$

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

- 6 years, 4 months ago

SOLUTION 26

Consider

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

Our integral is

$\displaystyle I'(1) = 2 \Gamma'(1)\beta(1) + 2 \Gamma(1) \beta'(1) = 2(-\gamma)(\pi/4) + 2 \frac \pi 4 \left[\gamma + 2 \ln 2 + 3 \ln \pi - 4 \ln \Gamma \frac 1 4 \right].$

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

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

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

which equals the stated result.

- 6 years, 4 months ago

PROBLEM 27

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

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

- 6 years, 4 months ago

Solution of Problem 27

Set $t=\tanh x$, we have

\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}

- 6 years, 4 months ago

Problem 28

Prove

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

- 6 years, 4 months ago

Solution 28

The integral equals

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

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

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

where

\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

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

Now using the identity

$\displaystyle \tan\frac{A+B}{2} = \frac{\sin A + \sin B}{\cos A + \cos B}$

gives

$\displaystyle J_\pm = \pm \frac 1 2 \frac{\sin 2}{\cosh 2 + \cos 2},$

which gives the desired result.

- 6 years, 4 months ago

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

- 6 years, 4 months ago

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

$\displaystyle \int_0^1 dx \frac{x^{a-1} - x^{-a}}{1-x} = \psi(1-a) - \psi(a) = \pi \cot \pi a,$

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

- 6 years, 4 months ago