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# 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
2 years, 4 months ago

## Comments

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Thanks for doing this. There is a lot to learn from these integration questions that you have shared. Staff · 2 years, 4 months ago

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Thank you for your help. You're too kind to me. I really appreciate it $$\quad$$ $$\ddot\smile$$ · 2 years, 4 months ago

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Problem 20

$$\displaystyle \int_0^1 \frac{\sinh ^{-1}(x)-\log \left[\left(\sqrt{2}-1\right) \sqrt{x}+1\right]}{x} \, dx = \log (2) \log \left(1+\sqrt{2}\right)-\frac{\pi ^2}{24}$$ · 2 years, 4 months ago

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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{align} \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{align}

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

\begin{align} \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{align}

Performing partial fractions decomposition we get

\begin{align} 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{align}

and

\begin{align} 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{align}

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{align} \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{align} and special values \begin{align} \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{align} 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 · 2 years, 4 months ago

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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. · 2 years, 4 months ago

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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! · 2 years, 4 months ago

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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 :) · 2 years, 4 months ago

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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.

· 2 years, 4 months ago

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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{align} \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{align} Consider $f(x)=\left\{ \begin{array}{l l} e^{i\omega x} & \quad \text{for}\ -\frac\pi2<x<\frac\pi2\\[12pt] 0 & \quad \text{otherwise} \end{array} \right.$ The Fourier transform of $$f(x)$$ is \begin{align} \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{align} Using $$(2)$$, then $$(1)$$ turns out to be \begin{align} \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{align} Now, let us express $$\dbinom{y}{z}$$ in term of beta function that can be related to $$(3)$$. \begin{align} \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{align} Using $$(4)$$, then $$(3)$$ turns out to be \begin{align} \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{align} · 2 years, 4 months ago

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OK, fine. I'll post high school integral problems from now. -_-" · 2 years, 4 months ago

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No Just keep posting those hard integrals, it's challenging but we learn a lot from it · 2 years, 4 months ago

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Your solution is valid only if $$v$$ is an integer, whereas the identity holds in general also. · 3 months, 1 week ago

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$$Problem\quad 23$$

Find $$\displaystyle \int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(tan(x))dx }$$ · 2 years, 4 months ago

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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. · 2 years, 4 months ago

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Sir can you elaborate I did'nt understood this one @Ruben Doornenbal · 2 years, 2 months ago

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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. · 2 years, 2 months ago

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@Ruben Doornenbal Can we have problem 24? · 2 years, 4 months ago

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PROBLEM 21 :

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

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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. · 2 years, 4 months ago

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Expecting a question from you @Ruben Doornenbal · 2 years, 4 months ago

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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. $$\ddot\smile$$ · 2 years, 4 months ago

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Yep, this will be good :) · 2 years, 4 months ago

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$$Problem\quad 30$$

Find $$\displaystyle \int _{ 1 }^{ \infty }{ \frac { x-\left\lfloor x \right\rfloor -0.5 }{ x } dx }$$ · 2 years, 4 months ago

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$I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} \int_{r}^{r+1} \frac{x-r-\frac{1}{2}}{x} dx$ $I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 1-(r+\frac{1}{2})\ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+1)\ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(\frac{r+1}{r}) + \ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(r+1) +(2r+2)\ln(r)+ \ln(\frac{r+1}{r})$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r)$ $2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r)$ $2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+2\ln(n!)-\ln(n)$ $2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+(2n+1)\ln(n)-2n+\ln(2\pi)-\ln(n)$ $2I = \ln(2\pi)-2$ $I =\frac{\ln(2\pi)}{2}-1$ · 2 years, 4 months ago

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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... · 2 years, 4 months ago

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Expecting a question from you @Tunk-Fey Ariawan , also please give a proof of your answer. · 2 years, 4 months ago

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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 :) . · 2 years, 4 months ago

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Exactly, you got it perfectly right. · 2 years, 4 months ago

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Problem 28

Prove

\begin{align} \int_{-\infty}^{\infty} \frac{\sinh 2x\cos 2x}{\sinh \pi x} \ dx = \frac{\sin 2}{\cos 2 + \cosh 2} \end{align} · 2 years, 4 months ago

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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{align} 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{align}

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. · 2 years, 4 months ago

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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. · 2 years, 4 months ago

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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 · 2 years, 4 months ago

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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! 👋(>‿◠) · 2 years, 4 months ago

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PROBLEM 27

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

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

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Solution of Problem 27

Set $$t=\tanh x$$, we have

\begin{align} \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{align} · 2 years, 4 months ago

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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. · 2 years, 4 months ago

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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. · 2 years, 4 months ago

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PROBLEM 24

$$\displaystyle \int_0^1 \operatorname{arcsech} x \operatorname{arcsin} xdx =\frac{\pi^2}{8} - \ln 2.$$ · 2 years, 4 months ago

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

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 :

$$I=\frac { { \pi }^{ 2 } }{ 8 } -ln(2)$$ · 2 years, 4 months ago

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Can you elaborate the first line please , it will be great to learn from you · 2 years, 2 months ago

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PROBLEM 22

This one is particularly beautiful, in my opinion.

$$\displaystyle \int_0^a \frac{x dx}{\cos x \cos(a-x)} = \frac{a}{\sin a} \ln \sec a.$$ · 2 years, 4 months ago

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

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

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

$$I=\frac { a }{ sin(a) } ln(sec(a))$$ · 2 years, 4 months ago

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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. · 2 years, 4 months ago

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

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There should be 3log(2) instead of 2log(2) · 2 years, 4 months ago

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PROBLEM 32

Find the closed-form expression of $$$\int_{-\infty}^{\infty} \frac{x\sin(x+\sin x)\,e^{\cos x}}{1+x^2}\,dx$$$ · 2 years, 4 months ago

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SOLUTION 32

We have

$$\displaystyle I = \Im \int_0^\infty \frac{x e^{ix + e^{ix}} dx}{1+x^2} = \Im \sum\limits_{k \geq 0} \frac{1}{k!} \int_0^\infty \frac{x e^{i(k+1)x} dx}{1+x^2} = \Im \sum\limits_{k \geq 0} \frac{1}{k!} \pi i e^{-(k+1)} = \pi e^{-1+e^{-1}}.$$

Here we made use of the fact that for positive $$a$$,

$$\displaystyle \int_0^\infty \frac{x e^{i a x} dx}{1+x^2} = \frac 1 i \frac{d}{da} \int_0^\infty \frac{e^{i a x} dx}{1+x^2} = \frac 1 i \frac{d}{da} \pi e^{-a} = i \pi e^{-a}.$$ · 2 years, 4 months ago

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I will post problem 33 when I get home. · 2 years, 4 months ago

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Post PROBLEM 33 here: Brilliant Integration Contest - Season 1 (Part 3). This note is already full. · 2 years, 4 months ago

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PROBLEM -31 Compute the integral $\int_{0}^{\pi} \ln(1-2a\cos(x)+a^2)dx$ , $$a > 1$$ · 2 years, 4 months ago

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SOLUTION OF PROBLEM 31

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

For a complete proof, you may refer to my answer on Math S.E. · 2 years, 4 months ago

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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. · 2 years, 4 months ago

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@Shivang Jindal Can you post your solutions for this integral when you have time? · 2 years, 3 months ago

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Please do share them for us here :) · 2 years, 4 months ago

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I am very interested in these solutions! · 2 years, 4 months ago

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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:

$$\displaystyle \int_0^1 x^2 \tan ^{-1}(x)^2 \, dx = \frac G 3 + \frac {1}{ 48} (4 - \pi)^2 - \frac {\pi}{12} \ln 2.$$ · 2 years, 4 months ago

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$$Solution\quad to\quad problem\quad 29$$

This one's easy.

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

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

$$\displaystyle I=\frac { { \pi }^{ 2 } }{ 48 } -\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \frac { ({ x }^{ 3 }+x-x){ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } dx }$$

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

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

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

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

Rest is trivial and it evaluates out to be :

$$J=\frac{\pi}{4}-\frac{1}{2}$$

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

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

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

Integration by parts $$dv=tan\theta d\theta , u=\theta$$

$$\displaystyle K=\theta ln(sec\theta )\overset { \frac { \pi }{ 4 } }{ \underset { 0 }{ | } } +\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cos\theta )d\theta }$$

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

Now $$\displaystyle M=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cos\theta )d\theta } =\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta } -\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(sin\theta )d\theta }$$

Adding both these forms we get :

$$\displaystyle M=\frac{1}{2}\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cot\theta )d\theta } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta }$$

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

Here I have two general results :

$$\displaystyle \frac { -\pi ln(2) }{ 2 } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta },\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cot\theta )d\theta } =G$$

Where $$G$$ is the catalan's constant,.

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

Using all these values we finally get $$I$$ as :

$$\large I=\frac{{\pi}^{2}}{48}+\frac{G}{3}+\frac{1}{3}-\frac{\pi}{6}-\frac{\pi ln2}{12}$$

$$\Rightarrow \large \boxed{I=\frac { { (4-\pi ) }^{ 2 } }{ 48 } +\frac { G }{ 3 } -\frac { \pi ln(2) }{ 12 } }$$ · 2 years, 4 months ago

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PROBLEM 17 :

Show that $\int_{-1}^{1}\frac{dx}{\sqrt[\Large3]{(1-x)(1+x)^2}}=\frac{2\pi}{\sqrt{3}}$ · 2 years, 4 months ago

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SOLUTION OF PROBLEM 17

Substitute $$x = 2t - 1$$ to get

$$I = \displaystyle{$$\int_0^1 \frac{dt}{\sqrt[3]{(1-t)t^2} } = B(1/3,2/3) = \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1)} = \frac{\pi}{\sin \pi/3} = \frac{2 \pi}{\sqrt 3}$$}$$ · 2 years, 4 months ago

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You may learn $$\LaTeX$$ around here from this tutorial note by @Daniel Liu . Anyway, don't forget to post your problem (PROBLEM 18) in new thread. Read the rules first. Thanks. · 2 years, 4 months ago

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PROBLEM 33

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

where $$q$$ is a parameter.

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

My own solution is kind of complicated, there is probably an easier way. · 2 years, 4 months ago

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SOLUTION OF PROBLEM 33 :

I am surprised that no one answers this question.

Okay, here is the solution. Setting $$t=x^2$$ and followed by partial fractions decomposition, we have \begin{align} \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{align} Setting $$x=\sqrt{q}\tan\theta$$ to $$I$$, we have \begin{align} 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{align} Differentiating $$J$$ w.r.t. $$q$$, we have \begin{align} 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{red}{\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{red}{\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{align} Now, to determine the value of $$C$$, let us evaluate $$J(0)$$. \begin{align} 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{align} Since $$J(0)=0$$, then $$C=0$$. Hence \begin{align} \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{align} It holds for $$q>0$$.

Although I know how to evaluate the second integral, i.e. use $$q=\frac{1}{2}$$, 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) · 2 years, 4 months ago

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Nice solution Anastasiya. I might need to think of a more difficult one for you next time ;). · 2 years, 4 months ago

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What do you mean by answering it violates the rules. · 2 years, 4 months ago

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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. · 2 years, 4 months ago

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Hint: the value of the first integral is

$$\displaystyle \pi \frac{2 \log \left(\sqrt{q}+1\right)-\sqrt{q}\log(4q) }{1-q}.$$ · 2 years, 4 months ago

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$$Problem\quad 25$$

Find $$\displaystyle \int _{ 0 }^{ 1 }{ ln(ln(\frac { 1 }{ x } ))dx }$$ · 2 years, 4 months ago

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

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

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

$\lim_{a\to1}\partial_a\,\Gamma(a)=\lim_{a\to1}\partial_a\,\int_0^\infty t^{a-1}e^{-t}\, dt=\lim_{a\to1}\psi(a)\Gamma(a)=-\gamma$

where $$\psi(x)$$ is the digamma function and $$\gamma$$ is Euler–Mascheroni constant. · 2 years, 4 months ago

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How can you prove that derivative of gamma function at one is equal to euler's mascheroni constant. @Anastasiya Romanova · 2 years, 4 months ago

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from definition of the digamma function we have

$\psi(x)=\frac{d}{dx}\ln\Gamma(x)=\frac{\Gamma'(x)}{\Gamma(x)}$

Hence

$\Gamma'(x)=\Gamma(x)\psi(x)$

and

$\Gamma'(1)=\psi(1)$

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

$\psi(n)=H_{n-1}-\gamma$

where $$H_{n-1}$$ is the harmonic number and by definition $$H_0=0$$ or you may refer to this and post on Math S.E. where I also contribute an answer there. · 2 years, 4 months ago

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$Problem\quad 19\quad :\\ Prove\quad that\quad :\\ \\ \int _{ \frac { -1 }{ \sqrt { 3 } } }^{ \frac { 1 }{ \sqrt { 3 } } }{ \frac { { x }^{ 4 } }{ 1-{ x }^{ 4 } } } \cos ^{ -1 }{ (\frac { -2x }{ 1+{ x }^{ 2 } } ) } .dx\quad =\quad \frac { \pi }{ 4 } \ln { (2+\sqrt { 3 } ) } \quad +\quad \frac { { \pi }^{ 2 } }{ 12 } \quad -\quad \frac { \pi }{ \sqrt { 3 } } \\ \\$ · 2 years, 4 months ago

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Solution to problem 19

We will integrate by parts to get rid of the arccos. The remaining integral will be trivial by symmetry. Observe that (at least for $$0 < x < 1$$)

$$\displaystyle \partial_x \cos^{-1} \left( \frac{2x}{1+x^2}\right) = \frac{2(1+x^2) - (2x)(2x)}{1+x^2} \frac{-1}{\sqrt{1 - \left(\frac{2x}{1+x^2} \right)^2}} = \frac{-2}{1+x^2}.$$

Also,

$$\displaystyle \frac{x^4}{1-x^4} = -1 + \frac{1}{2} \left[\frac{1}{1+x^2} + \frac{1}{1-x^2} \right].$$

Integrating by parts now yields

$$\displaystyle \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} dx \frac{x^{4}}{1-x^{4}}\cos^{-1}\left(\frac{2x}{1+x^{2}}\right) = \left( -x + \frac{1}{2} \tan^{-1} x + \frac{1}{2}\tanh^{-1} x \right) \cos^{-1}\left(\frac{2x}{1+x^{2}}\right) \left. \right\lvert _{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} - \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} dx \left( -x + \frac{1}{2} \tan^{-1} x + \frac{1}{2} \tanh^{-1} x \right) \frac{-2}{1+x^2} \\= \pi \left[\frac{\pi}{12} - \frac{1}{\sqrt 3} + \frac{1}{4} \log \left( 2 + \sqrt{3} \right) \right],$$ because the integrand is odd. Here we used the easy fact that $$\displaystyle 2 + \sqrt 3 = \frac{\sqrt 3 + 1}{\sqrt 3 - 1}.$$ · 2 years, 4 months ago

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