By comparison with the integral of $ \frac xae^{-x^{2}}$  Show that $\int_a^\infty e^{-x^{2}}dx≤ \frac 1{2a}e^{-a^{2}}   $ Given that $a&gt;0$.
66868
$X \sim \mathcal{N}(0,1)$, then to show that for $x &gt; 0$, $$ \mathbb{P}(X&gt;x) \leq \frac{\exp(-x^2/2)}{x \sqrt{2 \pi}} \&gt;. $$
80207
$ \int^\infty_0 (x^n)(e^{-x})  dx  $ and show that it is equal to $(n!)$ ?  I know that if you differentiate $x^n$  infinitely you get $n!$ but I don't know how to prove?
33935
Show that $$\int_{0}^{\infty}\frac{x\cos ax}{\sinh x}dx=\frac{\pi^2}{4} \operatorname{sech}^2\left(\frac{a\pi}{2}\right) $$   I think we can solve it by contour integration, but how? and it would be better if there were two ways, the first with real analysis and the second with complex analysis.
56366
I was reading an article in which it was stated that, with a change of variable, one could show that:  $$\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}$$  I tried with $t = 1 + \frac{1}{x}$ but that doesn't work out, especially because the lower bound doesn't become $0$.
49255
I read somewhere that   $$ \int_{0}^{\infty} \frac{x^3}{e^{x} - 1} = \frac{\pi^4}{15}$$  Does anyone see a way to prove this? My first idea was doing a contour integration and use the residue theorem, but that seems to be a lot of work so if anyone has a better idea I'd be glad to hear it.
105702
I know that $$\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$$ For substituting $u=2$ into $$\zeta(u)\Gamma(u)=\int^\infty_0\frac{x^{u-1}}{e^x-1}dx$$  However, I suspect that there is an easier proof, maybe by the use of complex analysis. I haven't learnt the  yet. All I know is the above formula and $\zeta(2)=\frac{\pi^2}{6}$. Can we can use the above integral to find out some of the $\zeta$'s value?
11444
I'm having a hard time proving the following $$\int_0^{\infty} \frac{x^8 - 4x^6 + 9x^4 - 5x^2 + 1}{x^{12} - 10 x^{10} + 37x^8 - 42x^6 + 26x^4 - 8x^2 + 1} \, dx = \frac{\pi}{2}.$$  Mathematica has no problem evaluating it while I haven't the slightest idea how to approach it. Of course, I would like to prove it without the use of a computer. Is this an explicit form of a special function I fail to recognize?
31614
How can I evaluate $$I=\int_{0}^{+\infty}\!e^{-ax^2-\frac b{x^2}}\,dx$$ for $a,b&gt;0$?  My methods:  Let $a,b &gt; 0$ and let $$I(b)=\int_{0}^{+\infty}e^{-ax^2-\frac b{x^2}}\,dx.$$ Then $$I'(b)=\int_{0}^{\infty}-\frac{1}{x^2}e^{-ax^2-\frac b{x^2}}\,dx.$$  What the other methods that can I use to evaluate it? Thank you.
92455
$$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$  I am not sure how to solve this question. Can anyone help me to approach this problem. Thanks.
12892
Proving that $$\sum_{n=0}^{\infty }\frac{1}{(2n)!!}=\sqrt{e}$$ Firstly, I tried to check the value with the exponential function at $x=.5$ but I found its terms not equal to the series terms.
88604
Essentially, all is in the title: Is it possible to compute the integral $$\int_1^{+ \infty} \frac{dx}{x^2(1+e^x)} \hspace{1cm} ?$$  I suspected some relation with a , but I was not able to find something explicit.
46433
I'm asked to show  $\int_0^{\infty} \frac{x\sin ax}{x^2+t^2}dx = \frac{\pi}{2}e^{-at}$ when $t,a &gt; 0$. I tried using integration by parts integrating $\frac{x}{x^2+t^2}$. But it seems that $\frac{x\sin ax}{x^2+t^2}$ has no antiderivative. What should I try?
47463
It is quite easy to show that (by using residue theorem) $$\int_0^{\infty} \frac{1}{1+x^n}dx = \frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)} $$ for $$n \ge 2$$  Is there any possibility to simplify $$\frac{2 \pi i^{1+2/n}}{n(e^{2 \pi i / n} - 1)}$$ or it is best result?  Thanks in advance!
120276
I was generalizing the following integral: $$\int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x \hspace{40pt} n\geq 0$$ We can start by noting that $\sin x =\dfrac{e^{ix}-e^{-ix}}{2i}$ and thus $\displaystyle \sin^{2n+1}x=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\sin(2r+1)x$. This implies $$\begin{aligned}\displaystyle \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x &amp;=\frac{(-1)^n}{2^{2n}}\sum_{r=0}^n (-1)^r \binom{2n+1}{r}\int_{0}^{\infty}\frac{\sin(2r+1)x}{x}\mathrm{d}x \\ &amp;=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n(-1)^r\binom{2n+1}{r} \\ &amp;=\frac{(-1)^n\pi}{2^{2n+1}}\sum_{r=0}^n\left((-1)^r\binom{2n}{r}-(-1)^{r-1}\binom{2n}{r-1}\right)\end{aligned}$$ Where I have used the well known  $\displaystyle \int_{0}^{\infty} \frac{\sin(2r+1)x}{x}\mathrm{d}x=\int_{0}^{\infty}\frac{\sin x}{x}\mathrm{d}x=\frac{\pi}{2}$ and the property of binomial coefficients that $\displaystyle \binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$ Since the above sum telescopes, we have $$\displaystyle \int_{0}^{\infty}\frac{\sin^{2n+1}x}{x}\mathrm{d}x=\frac{(-1)^n\pi}{2^{2n+1}}(-1)^n\binom{2n}{n}=\frac{\pi}{2^{2n+1}}\binom{2n}{n} ~\forall ~ n\in \mathbb{Z^{+}}$$ I would like to know other methods for evaluating this integral.
66782
I'm trying to find $$ \lim_{n \to \infty}   \int_1^a \dfrac{n}{1+x^n} \, dx  ,$$ where parameter $a&gt;1$. The limit appears to be $\ln 2$ regardless of the value of $a$. Some trick is probably needed here, but i don't see it.
11415
Prove the following using complex analysis   $$\tag{1}\int^\infty_0\frac{\ln(1+x)}{1+x^2}\,dx=\frac{\pi}{2}\ln(2)$$  I found this problem in Schaum's outlines of complex variables.   I thought that we could solve the problem using a similar method to evaluating   $$\int^\infty_0\frac{\ln(1+x^2)}{1+x^2}\,dx$$   Where we integrate the function  $$f=\frac{\log(1-iz)}{z^2+1}$$  around half a circle in the upper half plane and chosing an analytic branch of the logarithm in the region.   I spent so much time on (1) without success.
51644
I know that $\int{\frac{1}{x}}dx$ is simply $\ln{(x)}+c$ (-which is clearly unrelated to the problem but I just thought I would share anyway) but I am not sure how to approach $e^{x{^2}}$. Perhaps a substitution?