Today @integralsbot at twitter  an interesting integral relation for complicated integral that gives $\pi$ as a result. I don't know how this bot works or where does it take the relation it posts from, but some of the problems are really difficult to prove and interesting. Such as this example. $$\int_{0}^{\infty} \frac{\left(\frac{\pi  x}{2}-\log (x)\right)^3}{\left(x^2+1\right)^2 \left(\log^2(x)+\frac{\pi ^2}{4}\right)} = \pi \tag{1}\label{eq1}$$ I've tried to calculate this and it appears to be very difficult (for starters wolfram alpha can't calculate it). The plot of the function under the integral doesn't look very extraordinarily:  Using partial fraction decomposition the integral  to the following form: $$\int_{0}^{\infty} \left( \frac{3 \pi  x}{2 \left(x^2+1\right)^2}-\frac{\log (x)}{\left(x^2+1\right)^2}+\frac{\pi^3 x^3-6 \pi ^2 x^2 \log (x)-3 \pi ^3 x+2 \pi ^2 \log (x)}{2 \left(x^2+1\right)^2 \left(4\log ^2(x)+\pi ^2\right)} \right)$$ The first two terms of the integral can be  $$ \int_{0}^{\infty} \frac{3 \pi x}{2 \left(x^2+1\right)^2} = \frac{3 \pi}{4} $$ or  easily solved $$ \int_{0}^{\infty} -\frac{\log (x)}{\left(x^2+1\right)^2} = \frac{\pi}{4} $$ to yield: $$ \int_{0}^{\infty} \left( \frac{3 \pi  x}{2 \left(x^2+1\right)^2} - \frac{\log (x)}{\left(x^2+1\right)^2} \right) = \pi \tag{2}\label{eq2} .$$ This is interesting by itself, since plot of the integrated function has interesting shape with two inflection points of the first derivative, which I wouldn't expect to be equal to $\pi$ (of course the algebra clearly shows that it is)  What's however even more interesting is that, if I'm not mistaken, it follows from \eqref{eq1} and \eqref{eq2} that $$ \int_{0}^{\infty} \frac{\pi^3 x^3-6 \pi ^2 x^2 \log (x)-3 \pi ^3 x+2 \pi ^2 \log (x)}{2 \left(x^2+1\right)^2 \left(4\log ^2(x)+\pi ^2\right)} = 0 \tag{3}\label{eq3}.$$ This seems to be correct when checked with numerical integration (I've checked in relatively high precision of ~100 decimal digits) but seem to be very hard to prove. Plot of the function under the integral \eqref{eq3} doesn't appear (to me) to suggest any trivial solution:  The most obvious way to transform integral \eqref{eq3} is to expand the numerator which gives: $$  \int_{0}^{\infty} \left( -\frac{3 \pi ^2 x^2 \log (x)}{\left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}-\frac{3    \pi ^3 x}{2 \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}+\frac{\pi ^2 \log    (x)}{\left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)}+\frac{\pi ^3 x^3}{2    \left(x^2+1\right)^2 \left(4 \log ^2(x)+\pi ^2\right)} \right), $$ but the resulting integrals are also very difficult, perhaps too difficult to directly attack the problem in this way (I've tried to solve the second one, since it seems to have the simplest numerator, but didn't managed to get any results yet) and neither their numerical values (-1.93789..., 3.229820..., -0.322982..., -0.968946...) nor the plots indicate any obvious cancellations. Maybe there is some indirect way? Perhaps the integral \eqref{eq3} can be shown to be equal to $0$ without directly solving the integral at all? As I mentioned in the beginning of this post, I don't know where does this \eqref{eq1} relation come from but it seems to be true and interesting and must've been found somehow. Therefore I believe some solution of this problem exists. I post it as an interesting challenge/puzzle. Maybe someone knows some of the relations, can solve some of the unsolved integrals, sees clever substitution or can provide any interesting insight into this problem. So, to finally pose the question. How to prove/solve either \eqref{eq1} or \eqref{eq3}?
120960
Prove $$\int_0^\infty \frac{(\pi x - 2\log{x})^3}{\left(\log^2{x} + \frac{\pi^2}{4}\right)(1+x^2)^2} \text{d}x = 8\pi$$   I tried using a modified version of the integrand and an origin-indented semicircular contour on the positive real half of the complex plane.  Despite my best efforts, I am unable to retrieve the desired integral, as the negative factor from the negative imaginary axis is preventing me from being able to simplify the integrals.  On top of that, the residue doesn't come anywhere close to the exact result.  Are there better methods that I am missing?
115406
How to find      $$ I=\int_{0}^{1}{\ln^{2}\left(x\right)\ln^{2}\left(1 - x\right) \over 2 - x} \,{\rm d}x $$   My idea:   Let $x=1-t$, then     $$ I =\int_{0}^{1}{\ln^{2}\left(1 - x\right)\ln^{2}\left(x\right) \over 1 + x}\,{\rm d}x $$   I can't proceed any further. I can't remember that Math SE has this similar problem already posted.  this integral problem is from This Euler Sums  : $$\sum\limits_{n = 1}^\infty  {\frac{1}{{{n^2}}}} \left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{k}} } \right)\left( {\sum\limits_{k = 1}^n {\frac{{{{\left( { - 1} \right)}^{k - 1}}}}{{{k^2}}}} } \right) =  - \frac{{23}}{{1440}}{\pi ^4}\ln 2 + \frac{1}{{18}}{\pi ^2}{\ln ^3}2 - \frac{1}{{18}}{\ln ^5}2 - 4L{i_4}\left( {\frac{1}{2}} \right)\ln 2 - 8L{i_5}\left( {\frac{1}{2}} \right) + \frac{1}{{12}}{\pi ^2}\zeta \left( 3 \right) + \frac{{27}}{4}\zeta \left( 5 \right)$$  when I deal this series,then we must solve this integral .  Thank you.
88631
Problem: Show that $\int_{0}^{\infty}|\frac{ \sin x}{x}|=\infty $ .  Here is what i have tried :  I found out that for all $x&gt;0$ , $x-\frac{x^3}{6} &lt; \sin x &lt;x $  . So $1-\frac{x^2}{6}&lt;\frac{ \sin x}{x}&lt;1$ . So applying integration to both side from $0$ to $M$ we have $M-\frac{M^3}{18}&lt;\int_{0}^{M}|\frac{ \sin x}{x}|&lt; M$ . Now we send $M \rightarrow \infty$ and observe that left-hand side goes to infinity . Hence proved .  Please point out if there is anything wrong with my approach . I will also be delighted if could provide alternative solutions . Thank you .
87553
How to figure out the following integral? I have not been able to solve it from some time. $$\int_0^1\frac{x^{2} -1}{\log x}\,dx$$
96422
I came across the following integral, and I don't know how to solve it. $$ \int\log\left(x+\sqrt{x^2-1}\right)dx $$ I tried the "obvious" substitution of $x=\sec\theta$, which gives you: $$ \int\tan\theta\sec\theta\log\left(\tan\theta+\sec\theta\right)d\theta $$ However, this doesn't simplify it much, in the sense that I have no idea how to solve this now! Take out the common factor of $\sec$ from the log, or perhaps do $u=\tan\theta+\sec\theta$, but both of these lead to dead ends (at least for me).  In case you are wondering, Wolfram|Alpha claims the answer is: $$x\log\left(x+\sqrt{x^2-1}\right)-\sqrt{x^2-1}+c$$
4264
How to compute the following integral?  $$\int\log(\sin x)\,dx$$    Motivation: Since $\log(\sin x)'=\cot x$, the antiderivative $\int\log(\sin x)\,dx$ has the nice property $F''(x)=\cot x$. Can we find $F$ explicitly? Failing that, can we find the definite integral over one of intervals where $\log (\sin x)$ is defined?
46239
This integral popped up recently  $$\int_{-\infty}^{\infty} \frac{e^{2x}-e^x}{x (1+e^{2x})(1+e^{x})}\mathrm{d}x = \log 2$$  A solution using both real and complex analysis is welcome. I tried rewriting it using symmetry, and then the series expansion of $1/(1+e^{nx})$, however this did not quite make it.   \begin{align*} I &amp; = 2\int_{0}^{\infty} \frac{e^{2x}-e^x}{x e^{3x}(1+e^{-2x})(1+e^{-x})}\mathrm{d}x \\ &amp; = 2 \int_{0}^{\infty}\frac{e^{-2x} -e^{-x}}{x} \left(\sum_{n=0}^\infty  (-1)^n e^{-n x}\right)\left(\sum_{m=0}^\infty (-1)^me^{-2mx}\right) \mathrm{d}x  \end{align*} The first part reminds me of a Frullani integral (it evaluates to $\log 2$). However I am unsure if this is the correct path, any help would be appreciated. =)
42380
I have in trouble for evaluating following integral  $$\int_0^{\infty}  \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$$  It seems really easy, but I don't know how to handle it at all. (The results are well known, here I tried to evaluate it but I failed)  I tried to use the relation   $$\sqrt{1+x^{4}}-x^{2}=\frac{1}{\sqrt{1+x^{4}}+x^{2}}$$   but I couldn't find the desired results.
77661
If $I_n=\int _0^{\pi }\sin^{2n}\theta \:d\theta $, show that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $  Hence calculate $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$  I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question.
111619
I need help evaluating the following integral $$\int_{0}^{\infty} \frac{\sin^4 x}{x^2}\,dx$$ which should probably be equal to $\frac{\pi}{4}$  Using some trigonometric manipulations I got $\frac{3}{8} - \frac{\cos{2x}}{2} + \frac{\cos{4x}}{8}$ which using integration by parts doesn't lead me to anything pretty.  Update: Not sure if I should post this as a separate question but getting explanation why $\int_{0}^{\infty} \frac{\sin{ax}}{x} = \frac{\pi}{2}$ for a positive integer $a$ could help me solve this question.
75595
I have been attempting to solve this integral for several days in preparation for an exam and keep reaching dead ends.  I(a,b) = $\int_0^{\pi/2} \ln(a^{2}\cos^2(x)+b^{2}\sin^2(x)) dx$, where $a,b&gt;0$  My best approach so far is defining:  $F(x) = I(x, b)$  Then, I know:  $F(b) = I(b, b) = \pi/2 * \ln(b^2)$  Then I am now attempting to find $F'(x)$ using differentiation under the integral sign, and integrating from b to a.  This is where I am stuck - no matter what I do, I cannot solve the integral. Am I in the right direction or should this be solved in a different manner? I also considered something elliptical due to the acos, bsin in the original integral, yet it has also reached a dead end.
24552
I am solving the following integral: $$\int_0^\infty \frac{x}{1+x^3}dx$$ I need to solve it using integration and residue theorems. I tried to convert it to complex function, for example $\frac{z}{1+z^3}$, but it is not even, so it is difficult to find the curve for integration.   Any hints are welcome!  Thank you!
24500
Show that $\displaystyle{\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2} = \frac{\pi}{2}\log \left(\sqrt{2\pi} \Gamma\left(\frac{3}{4}\right) / \Gamma\left(\frac{1}{4}\right)\right)}$  This question was posted as part of this question:     I cannot think of a change of variable nor other integrating methods. Maybe there is a known method that I am missing.
85817
How do I solve this? $$\int_0^\infty\frac{\ln(1+{x^2})}{1+{x^2}}$$ I know the answer is $\pi\ln2$.
39730
This integral below $$ I:=\int_0^\infty \log^2 x\frac{1+x^2}{1+x^4}dx=\frac{3 \pi^3}{16 \sqrt 2} $$ is what I am trying to prove.  Thanks.  We can not expand the denominator as a series since the domain of integration is for $x\in [0,\infty)$.  Next I wrote $$ I=\int_0^\infty \log^2 x \frac{1+x^4-x^4+x^2}{1+x^4}dx=\int_0^\infty \log^2x \left(\frac{1+x^4}{1+x^4}+\frac{x^2-x^4}{1+x^4}\right)dx=\\ \int_0^\infty \log^2 x \, dx+\int_0^\infty \log^2 x \frac{x^2}{1+x^4}dx-\int_0^\infty \log^2 x \frac{x^4}{1+x^4}dx, $$ however only the middle integral is convergent.  I am not sure how to go about solving this problem.  Thank you
107304
The problem is to show  $$\int_0^\infty \frac{e^{-x}-e^{-xt}}{x}dx = \ln(t),$$  for $t \gt 0$.  I'm pretty stuck. I thought about integration by parts and couldn't get anywhere with the integrand in its current form. I tried a substitution $u=e^{-x}$ and came to a new integral (hopefully after no mistakes)  $$ \int_0^1 \frac{u^{t-1}-1}{\log(u)}du, $$  but this doesn't seem to help either. I hope I could have a hint in the right direction... I really want to solve most of it by myself.  Thanks a lot!
123571
I want to show that $\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0$, but is there a faster method than finding the contour and doing all computations?  Otherwise my idea is to do the substitution $x=e^t$, integral than changes to $\int _{-\infty}^{\infty}\frac{t e^t}{1+e^{2t}}dt$. Next step is to take the contour $-r,r,r+i\pi,-r+i\pi$ and integrate over it...