I need to express the following improper integral as a double integral of $x$ and $y$ and then, using polar coordinates, evaluate it.  $$I=\int_{-\infty}^{\infty}e^{-x^2}dx$$  Plotting it, we find a Gaussian centered at $x=0$ which tends to infinity to both sides. We can easily express it as a double integral :  $$I=\int_{0}^{1}\int_{-\infty}^{\infty}e^{-x^2}dxdy$$  Evaluating both using Wolfram Alpha gives $\sqrt{\pi}$, so it converges.  I know that $x=rcos(\theta)$ and that $dxdy=rdrd\theta$, but substituing this in the above integral and evaluating $\theta$ from $0$ to $2\pi$ and $r$ from $0$ to $\infty$ doesn't yield the correct answer. What's wrong here?  Thanks a lot !
2645
How to prove  $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$
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.
20679
We know that the integral $$ I = \int_{-\infty}^{\infty}  e^{-x^2}\ \mathrm{d} x $$ can be calculated by first squaring it and then treat it as a $2$-dimensional integral in the plane and integrate it in polar coordinates. Are there any other ways to calculate it?  I know that we may use the relation $$\Gamma(x)\, \Gamma(1-x) = \frac{\pi}{\sin{\pi x}}$$ but this, in effect, is still taking the square. Well, after I write down the above text, I figure that maybe there is no way to calculate it without squaring, since, after all, the result contains a square root, and it seems no elementary function can &quot;naturally&quot; produce a square root of $\pi$ starting from natural numbers (though I don't know how to describe this more concretely; you are also welcome to comment on this point).  Nevertheless I still post this question in case there are some other ideas. EDIT: the Fourier transformation method at  appears kind of cheat to me, since the very proof of the Fourier transformation formula actually makes use of the value of the Gauss integral (at least in this wiki page ). Thank you.
8145
I have been self-studying CA and find it very interesting. So, working through problems in a book I have, I ran across  $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{2\pi x}-1}dx=\frac{1}{4}\coth(a/2)-\frac{1}{2a}$$  and $$\int_{0}^{\infty}\frac{\sin(ax)}{e^{x}+1}dx=\frac{1}{2a}-\frac{\pi}{2\sinh(\pi a)}$$  For the former, I wrote it as $\frac{e^{aiz}}{e^{2\pi z}-1}$ and used a rectangular contour with vertices $0, \;\ R. \;\ R+i, \;\ i$  $e^{2\pi z}-1$ has poles at $ni$. Of these, I think $i$ only lies within the contour.   Unless I am in error, I then calculated the residue at $i$ to be $\frac{e^{-a}}{2\pi}$  So, $2\pi i(\frac{e^{-a}}{2\pi})=ie^{-a}$  Now, where I get hung up is setting up the integrals around the contour. Two of which should tend to 0 as $R\to \infty$.  Here is what I done.  $$\int_{0}^{R}\frac{e^{iax}}{e^{2\pi x}-1}dx+\int_{0}^{\infty}\frac{e^{ai(R+iy)}}{e^{2\pi (R+iy)}-1}idy+\int_{R}^{0}\frac{e^{ai(x+i)}}{e^{2\pi (x+i)}-1}dx+\int_{\infty}^{0}\frac{e^{ai(iy)}}{e^{2\pi iy}-1}dy=-\sinh(a)$$  I am unsure of the limits on the second and fourth integrals.  I am not so sure this is correct. The second and fourth ones, which represent the vertical sides,  should tend to 0 as $R\to\infty$. I hope :).   This than gave me $(1-e^{-a})\int\frac{e^{iax}}{e^{2\pi x}-1}dx=-\sinh(a)$.  Which does not look correct. I did manage to solve this using series and $\pi csc(\pi z)$, but the contour I am unsure of.   Can someone lend a hand here?. Any advice on either would be appreciated.   For the other one, I only posted it because it looks similar, but I believe is actually more involved and 'tougher' if you will. If I can get one, perhaps I can manage to evaluate the other.  With that one, I think the same rectangle with the same vertices can be used, but $\pi i$ would have to be avoided. Perhaps a Principal Value in there somewhere. But, I was told to use vertices $0, \;\ R, \;\ R+2\pi i, \;\ 2\pi i$  Since I am relatively new to CA, setting up those integrals is the confusing part.    I can do easier types of contour integrals, but want to learn more about these more challenging ones.   Thanks for any assistance.
75155
So I came across this question on brilliant.org and didn't know how to go about it: $$\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $$  I tried to complete the squares in the denominator and then use a trigonometric substitution (with tan) but I didn't get anything beyond that.   How do I do it? Do we need to differentiate under the integral sign here?
52340
I have tried the $u$ substitution yet I could not move forwards on this problem. Could you calculate in closed-form this integral? $$\int_{0}^{\pi/2}\bigg\{\csc(x)\bigg\}\mathrm{dx}$$
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?
123246
How can I calculate $$\int_0^\infty\frac{\sin x}xdx$$ by integration of a suitable function along the following paths:  where $R$ and $\varepsilon$ are the radius of the shown outer and inner semicircle, respectively. This seems to be quite interesting. How would one do that?
21986
I want to evaluate the integral   $$\int_{-\infty}^\infty \frac{\sin^2(x)}{x^2}e^{i t x} dx$$   for all $t \in \mathbb{R}$. I would preferably do it using the tools of complex analysis, but since I haven't found any other thread dealing with this integral at all so far, I'd be open to any approach.  I must admit that I haven't really found a way to start. The function $f(z) := \frac{\sin^2(z)}{z^2}e^{i t z}$ has an isolated singularity (a removable singularity, with continuation $f(0) = 1$, to be precice) at $z = 0$. I think a clever way would be to integrate over the edge of rectangle, semicircle or "slice of cake" (the edges of which we then would let go against $-∞$ and $∞$), then maybe somehow apply the Cauchy theorem or Residue theorem. But I don't know how to concretely do that, or how to even get started with that.  On a more general note, I'm having trouble understanding how to transform a "normal" integral over a real interval into one where we integrate over the edges of a geometric shape; and I also don't know, based on which pattern one can see with geometric shape fits best. Are there any tips or hints on how to get more intuition for that? (Apart from, obviously, dealing with these integrals regularly, what I'm trying to do but where I'm still failing a lot of times.)  I also noticed that the above integrals looks strikingly similar to the Fourier transformed of $f$ (or, even more similar to the formula for the inverse transformed). So maybe this could help us with this integral? I'm not really sure though.
47738
While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.      $$\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.$$
68926
I want to find the following integral: $$\int_{0}^{\infty} \frac{e^{3x}-e^x}{x(e^{3x}+1)(e^x+1)} dx$$ I have been trying to apply simple substitutions like $u=e^x$, to simplify the integral without any luck. I simplified it to  $$\int_{0}^{\infty} \frac{1}{x(e^x+1)}-\frac{1}{x(e^{3x}+1)}dx$$  which seems to have increased my problems.  Wolfram Alpha gives the answer to be 0.5493... and the graph looks like a Gaussian distribution.  Somehow, I feel that this integral is very non-trivial due to the 'troublesome' $x$ in the denominator.
33837
Improper integral $\sin(x)/x $ converges absolutely, conditionaly or diverges?  We have that $$\int_1^{\infty}\frac{\sin x}{x}dx$$ $$u=\frac{1}{x}$$ $$du=-\frac{1}{x^2}dx$$ $$dv=\sin xdx$$ $$v=-\cos x$$ $$\int_1^{\infty}\frac{\sin x}{x}dx=-\cos(1)-\int_1^{\infty}\frac{\cos x}{x^2}dx$$ $$\int_1^{\infty}\frac{\cos x}{x^2}dx&lt;\int_1^{\infty}\frac{1}{x^2}dx$$ So it converges. Now I need to find out if $\int_1^{\infty}|\frac{\sin x}{x}|dx$ converges or diverges.
111805
This is a not mandatory exercise I am struggling with from my Analysis II Class, at the very end of it I am supposed to compute $$\int_{-\infty}^\infty e^{-t^2}dt \tag{*}$$ The most famous way to do that is switch to polar coordinates but I am not supposed to do it that way, one of my tutors even called it a 'dirty trick'.  Instead the exercise gives us the following two Integrals to work with:      $$F(x)= \int_0^1 \frac{e^{-x^2(1+t^2)}}{1+t^2}dt \text{ and } G(x)= \left( \int_0^x e^{-t^2}dt\right)^2, \ \forall x \in \mathbb{R} $$   I yet fail to see how these two integrals relate to (*) but I can naively follow the instructions given by exercise sheet and hope that the result will follow:     Problem: Show that $F,G$ are of Class $C^1(\mathbb{R})$, compute $F',G'$ for all $x \in \mathbb{R}$ and show that $F+G$ is constant   My approach: There isn't much to do really rather than to quote some Lemmas, clearly the parametric integral $F$ is of class $C^1$ because $f(x,t): \mathbb{R} \times [0,1] \to \mathbb{R}$ given by $f(x,t)=e^{-x^2(1+t^2)}/(1+t^2)$ is of class $C^1$, I can therefore switch the partial derivative with the integral and obtain that  $$F'(x) =-2 x \int_0^1e^{-x^2(1+t^2)}dt$$ I doubt that I can further simplify this integral.  With a little help of the fundamental theorem of Calculus one obtains that $$G'(x)= 2 \left( \int_0^x e^{-t^2}dt\right)e^{-x^2} $$    At this point I hope that the exercise (or my computation) is flawed, at least the limit points of the integrals because I wouldn't know how to show that $F'+G'=0$ for all $x\in \mathbb{R}$ to conclude that $F+G$ is constant.   The only sense I could make out of $F'$ would be if I integrated it with respect to $x$ rather to $t$. Are there any further simplifications I can do?
53688
I'm trying to solve this integral: $$\int_{-\infty}^\infty e^{-x^2-x}dx$$ WolframAlpha shows this to be approximately $2.27588$. I tried to solve this by integration by parts, but I just couldn't get there. I'd be glad if someone could show me how to do it.   I've included my attempt at the problem below:  Note $\int_{-\infty}^\infty e^{-x^2-x} = \int_{-\infty}^\infty e^{-x^2}e^{-x}dx$. Then we can integrate by parts. Let $u(x) = e^{-x^2}$. Then $u'(x) = -2xe^{-x}$. Let $v'(x) = e^{-x}$. Then $v(x) = -e^{-x}$.  Then $u(x)v'(x) = u(x)v(x) - \int v(x)u'(x)dx$, i.e. $$\int_{-\infty}^\infty e^{-x^2}e^{-x}dx = -e^{-x^2}e^{-x} - \int e^{-x}2xe^{-x}dx = -e^{-x^2-x} - 2\int e^{-2x}xdx$$  Then we integrate $\int e^{-2x}xdx$ by parts. We pick $u(x) = x$, $u'(x) = dx$, $v'(x) = e^{-2x}$, and $v(x) = \int e^{-2x} dx = \frac {-1}{2} e^{-2x}$ by u-substitution. Skipping some steps, it follows that  $$\int e^{-2x}xdx = -\frac{1}{4}(e^{-2x})(2x+1)$$ Then  $$ -e^{-x^2-x} - 2\int e^{-2x}xdx = -e^{-x^2-x} + \frac{1}{2}(e^{-2x})(2x+1)$$ Which, when evaluated numerically, doesn't yield the desired result. So there's a problem here. Maybe someone else knows how to do this.
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 ?
10157
I have stumbled onto an interesting integral$$\int_0^\infty \sin(x)\sin\left(\frac{1}{x}\right)dx$$ which I noticed graphically that it appears to be $1$, but I have no idea on how to evaluate it. Maybe it could be done with the use of Bessel Functions? Any help is appreciated.
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. =)