$$\int_0^\infty {(\ln x)^2 \over {1+x^2}} dx$$ I need to evaluate the above integral using  to obtain the answer $\dfrac{\pi^3}8$:  However the only example involving rectangular contours I've seen in the textbook or in the many Youtube videos I've searched is the following: $$\int_{-\infty}^{\infty} {{e^{ax} \over {1+e^x}}dx}$$ where $a$ is a constant between $0$ and $1$. The hint is to use the substitution $x=e^t$ so I assume I'm suppose to turn the first integral into the form of the second, but then I end up with a $t^2$ I don't know what to do with. I am aware that if I solve the first integral from $-\infty$ to $\infty$ I just divide that answer by $2$ to get the answer from $0$ to $\infty$ so I guess the answer I'm really looking for is $\dfrac{\pi^3}4$.
109259
George Arfken's book: Mathematical Methods for Physicists has the following problem in a chapter on contour integration:  $\displaystyle \int_0^{\infty} \dfrac{(\log x)^2}{x^2 + 1} dx$.   Their suggestion is to make the substitution $x \rightarrow z=e^t$. I'm not sure what they meant by this, but I tried making the substitution $x = e^t$, which turns the integral into:  $\displaystyle\int_{-\infty}^{\infty} \dfrac{t^2 e^t}{1+e^{2t}} dt$.   The hint is then to take the contour from -R to R, to R+$\pi i$, to -R + $\pi i$, to -R. Since this has a pole at $t = \pi i/2$, a Laurent series expansion about this point gives the residue as $i \pi^2/8$, so the contour integral equals $-\pi^3/4$.  I've been able to show that the integral along R to R + $\pi i$ and along -R + $\pi i$ to -R goes to zero by the ML inequality - the denominator grows exponentially but the numerator quadratically.  But at this point, I'm a bit lost as to what to do with the integral over Im $t = \pi$. Any help? The book gives the answer as $\pi^3 /8$.
94619
How can I compute $$\int_0^\infty\frac{1}{(1+x^{2015})(1+x^2)}\quad?$$    My attempt: Looking at the limits of the integration I see that we should induce some $\tan^{-1}(x)$ so if we put $\infty$ we would get something like $\frac{\pi}{2}$ . But I am not sure how to proceed . Partial fractions don't yield good integral.
47260
I'm searching a closed form for the integral of the form :  $$ \int_{0}^{\infty}e^{-ax}e^{-x^\frac{\alpha}{2}}dx $$  especially for $\alpha=8$  After several attempts, I find that the general form can be written on the basis of hypergeometric functions. what do you think ?
39954
Calculate  $$\int^{\infty}_{0}\frac{\ln^2(x)}{(1-x^2)^2}dx$$   I have tried to put $\displaystyle x=\frac{1}{t}$ and $\displaystyle dx=-\frac{1}{t^2}dt$  $$ \int^{\infty}_{0}\frac{t^2\ln^2(t)}{(t^2-1)^2}dt$$  $$\frac{1}{2}\int^{\infty}_{0}t\ln^2(t)\frac{2t}{(t^2-1)^2}dt$$  $$ \frac{1}{2}\bigg[-t\ln^2(t)\frac{1}{t^2-1}+\int^{\infty}_{0}\frac{\ln^2(t)}{t^2-1}+2\int^{\infty}_{0}\frac{\ln(t)}{t^2-1}dt\bigg]$$  How can I solve it?
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.
17907
Most of us are aware of the classic Gaussian Integral   $$\int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$  I would be interested in evaluating the similar sum  $$\sum_{x=0}^\infty e^{-x^2}$$  Now, because $\exp(-\lfloor x \rfloor^2) \ge \exp(-x)$, we find   $$\sum_{x=0}^\infty e^{-x^2}= \int_0^\infty e^{-\lfloor x \rfloor^2}\, dx \ge \int_0^\infty e^{-x^2}\, dx=\frac{\sqrt{\pi}}{2}$$  Does a closed form for this sum exist?  If so, what would it be?  I would be very interested in how a closed form would be found for this function.
63223
Given the integral $\int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})} dx$  How can I evaluate it? Thanks!
44933
What contour and integrand do we use to evaluate $$ \int_0^\infty \frac{x^3}{e^x-1} dx $$  Or is this going to need some other technique?
20193
Find $\displaystyle \int_{0}^{\infty} \dfrac{\ln(x)}{1+x^2}dx$.   How should I change the limits of integration to evaluate this?
106434
$$I=\int_1^\infty \frac{1}{x(x^2+1)}\ dx$$    I tried to use partial fractions, but am unsure why I cann't evaluate it using partial fractions. Would appreciate any explanation about which step is incorrect.    $$\frac{1}{x(x^2+1)} = \frac{1}{2} \left(\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i}\right)$$ So   $$I = \frac{1}{2}\int_1^\infty\frac{1}{x}-\frac{1}{x+i} -\frac{1}{x-i} \ dx \\ = \frac{1}{2}\left[\log |x| - \log(|x+i|) - \log(|x-i|)\right]_1^\infty   $$ which evaluates to be $\infty$.
74926
I have seen the Fresnel integral  $$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$  evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral.  I was wondering, however, does anyone know a real analysis method to evaluate this integral?
26610
I'm trying to compute $\displaystyle \int_{-\infty}^\infty \frac{e^{ax}}{1+e^x}dx$, $(0&lt;a&lt;1)$ Let $f$ denote the integrand.  I'm using the rectangular contour given by the following curves: $c_1: z(t) = R+it, t \in [0, 2\pi]$ $c_2: z(t) = -t+2\pi i, t \in [-R, R]$ $c_3: z(t) = -R + i (2\pi - t), t \in [0, 2\pi]$ $c_4: z(t) = t, t \in [-R, R]$    There is one singularity within the contour, at $z = \pi i$. Expanding out the denominator as a power series shows that it's a simple pole, and allows us to evaluate the residue as $\displaystyle \lim_{z \rightarrow \pi i} f(z)(z-\pi i) = - e^{a \pi i}$    This is computed by expanding $1+e^z$ as a Taylor series around $\pi i$. The first coefficient will be 0, and the second will be $-1$. The rest will have orders of $(z - \pi i)$ greater than 1, and will thus vanish when we take the limit.  So the integral over the entire contour is $- 2\pi i e^{a \pi i}$    An easy enough estimate on the $c_1$ shows that the integral vanishes as $R \rightarrow \infty$. With a variable change, c_3 is the same as c_1 and also vanishes. $c_4$ becomes the integral we want when we take a limit. $c_2$ becomes $c_4$ with a constant:    \begin{align*} \int_{c_2} f(z)dz &amp;= \int_{-R}^{R} \frac{e^{-at}e^{a 2\pi i}}{1+e^{-t}e^{2\pi i}}dt \\ &amp;= e^{a 2 \pi i}\int_{-R}^{R} \frac{e^{-at}}{1+e^{-t}}dt \\ &amp;=e^{a 2 \pi i}\int_{R}^{-R} - \frac{e^{au}}{1+e^{u}}du \ \ \ (u = -t, du = -dt)   \\ &amp;= e^{a 2 \pi i}\int_{-R}^{R} \frac{e^{au}}{1+e^{u}}du \\ &amp;= e^{a 2 \pi i} I(R) \end{align*}    Where $I(R)$ is the line integral over $c_4$. Putting it all together and taking the limit gives us $\displaystyle \lim_{R \rightarrow \infty} I(R) = \frac{- 2\pi i e^{a \pi i}}{(1 + e^{a 2 \pi i}) }$    But this can't be the value of the integral, because it's a real-valued function integrated over $R$. I can't figure out where I'm going wrong. Note that I've avoided posting all the details of my solution since this is from a current problem set for a class on complex analysis.
69306
Evaluate the integral,   $$ \int_{0}^{1} \ln(x)\ln(1-x)\,dx$$   I solved this problem, by writing power series and then calculating the series and found the answer to be $ 2 -\zeta(2) $, but I don't think that it is best solution to this problem.   I want to know if it can be solved by any other nice/elegant method.
30836
Calculate   $$\frac{\int_0^\pi x^3\ln(\sin x)\,dx}{\int_0^\pi x^2\ln(\sqrt{2}(\sin x))\,dx}$$   In this problem, I'm unable to understand how to start.  I tried applying integration by parts but I couldn't solve it. I also tried the various properties of definite integration but they were of no use. Maybe applying integration by parts (or DI method) successively may work but it leads to a form of $\frac{\infty}{\infty}$.
116685
Evaluate  $$\int_0^\infty \frac{\log x \; dx}{x^{2} + 2x + 2}$$  by integrating a branch of $(\log z)^{2}/(z^{2} + 2z +2)$ along a keyhole contour.  The thing I have trouble with is why I should be examining the square of the log - I guess it has something to do with ln x in fact NOT being the real part of log z, since log z is (or can be) defined on the entire negative real axis and ln x can't. But looking at the square would definitely not have been my first plan of attack :(
56233
How to find  $$ \int_0^\infty \dfrac x{1+e^x}\ dx=\ ...? $$  I don't know where should I start with. The correct answer from my textbook is $\frac{\pi^2}{12}$. This is my homework with 10 questions but I can only answer 9 questions, this one I'm stuck. Can anyone help me? Thanks.
6312
Another exam problem I'm looking at is to evaluate the following integral. $$ \int_{-\infty}^{\infty} \frac{x\hspace{-0.04 in}\cdot\hspace{-0.04 in}\sin(x)}{x^{\hspace{.02 in}2}+1} dx $$ This is a complex analysis exam, so the solution probably involves contours. $\:$ Since the integrand is an even function, one could potentially simplify by changing one endpoint to $0$. However, I have no idea how to make a contour work since the absolute value of the integrand grows exponentially away from the real axis. What does one need to do to evaluate that integral?