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I just stumbled across the fact that $\int_{-\infty}^{+\infty}{e^{-x^2}dx}=\sqrt{\pi}$. This intrigued my already-existing interest in integrals. It made me wonder, are there other integrals with crazy results?

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    $\begingroup$ Could you please be more specific with what kind of integrals you're looking for? $\endgroup$
    – teadawg1337
    Dec 8, 2014 at 21:09
  • $\begingroup$ @teadawg1337 , integrals with unexpected results or integrals with just overall interesting results. $\endgroup$
    – Kurtbusch
    Dec 8, 2014 at 21:10
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    $\begingroup$ You should look for a course in complex analysis! $\endgroup$
    – Raclette
    Dec 8, 2014 at 21:10
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    $\begingroup$ See: the highly voted questions with the relevant tag(s). $\endgroup$
    – user147263
    Dec 8, 2014 at 21:34
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    $\begingroup$ If you mean "other integrals where $\pi$ shows up for no apparent reason" - yes, definitely, that's part of $\pi$'s job. $\endgroup$
    – Milo Brandt
    Dec 8, 2014 at 22:58

9 Answers 9

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One I just evaluated

$$\int_{-\infty}^{\infty} \frac{dx}{(e^x+x+1)^2+\pi^2} = \frac{2}{3}$$

or the ultimate insanity...

$$\int_{-1}^1\frac{dx}{x}\sqrt{\frac{1+x}{1-x}}\log\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) = 4 \pi \operatorname{arccot}{\sqrt{\phi}} $$

where $\phi=(1+\sqrt{5})/2$ is the golden ratio.

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This one, which is another version of your integral$$ \int_{-\infty}^{+\infty}e^{-\left(x+ \tan x \right)^2} \mathrm{d}x=\sqrt{\pi},$$ or this strange family $$ \begin{align} \displaystyle \int_0^{1} \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots + \cfrac 1 { 1 + \psi (\left\{1/x\right\}+1)}}}} \mathrm{d}x & = \dfrac{F_{n-1}}{F_{n}} - \dfrac{(-1)^{n}}{F_{n}^2} \ln \left( \dfrac{F_{n+2}-F_{n}\gamma}{F_{n+1}-F_{n}\gamma} \right),\\\\ \end{align} $$ where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$, $\gamma$ is the Euler constant, $F_{n}$ are the Fibonacci numbers, $\psi:=\Gamma'/\Gamma$ is the digamma function and where the continued fraction has $n$ horizontal bars.

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    $\begingroup$ Okay. This family what you've created. Not bad. +1. $\endgroup$
    – user153012
    Dec 9, 2014 at 1:30
  • $\begingroup$ @user153012 Thank you. $\endgroup$
    – Olivier Oloa
    Dec 9, 2014 at 20:08
  • $\begingroup$ Could you share more on how to evaluate the interesting family of integrals that you mentioned? $\endgroup$
    – Lucian
    Dec 9, 2014 at 22:08
  • $\begingroup$ @Lucian Yes. Maybe, you could ask a related question so I would have more room than the one for this comment? Thanks. $\endgroup$
    – Olivier Oloa
    Dec 9, 2014 at 22:23
  • $\begingroup$ Actually, the site-rules explicitly allow and even encourage users to ask-and-answer their own questions. $\endgroup$
    – Lucian
    Dec 9, 2014 at 22:28
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${\tt\mbox{Sophomore's Dream}}$: Discovered by Johann Bernoulli in 1697 !!!.

$$ \int_{0}^{1}x^{-x}=\sum_{n\ =\ 1}^{\infty}n^{-n} $$

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$\displaystyle \int_0^{+\infty}\dfrac{\sin x}{x}dx=\dfrac{\pi}{2}$

$\displaystyle \int_0^{+\infty}\cos(x^2)dx=\dfrac{1}{2}\sqrt{\dfrac{\pi}{2}}$

Math.stackexchange is full of such beautiful formulae.

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  • $\begingroup$ but there seems to be room for more... $\endgroup$
    – Jaume Oliver Lafont
    Feb 15, 2016 at 11:17
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$$\large\int_{-\infty}^\infty\, \frac{1}{1+(x+\tan x)^2}\, dx= \pi $$

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$$\begin{align} \int_0^\infty\exp\Big(-\sqrt[n]x\Big)dx&=n! \\ \\ \int_0^1\Big(1-\sqrt[n]x\Big)^mdx&={m+n\choose n}^{-1}={m+n\choose m}^{-1} \\ \\ \int_0^1\ln\Big(1-\sqrt[n]x\Big)dx&=-H_n \\ \\ \lim_{n\to\infty}n\Big(1-\sqrt[n]x\Big)&=-\ln x \end{align}$$

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$$\int_{\mathbf{R}}\frac{dx}{1+x^2}=\pi$$ and $$\frac{22}{7} - \pi = \int_0^1 \frac{(x-x^2)^4}{1+x^2}dx.$$

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    $\begingroup$ Actually, $\int_{\mathbb R} \frac{e^{itx}}{1+x^2} dx = e^{-|t|}\pi$ for any $t\in \mathbb R$. Note that the integrand is complex but the answer is real.. $\endgroup$
    – Raclette
    Dec 8, 2014 at 21:15
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$$\int_0^{\infty}\left(\frac{3}{4}\right)^{\lfloor{x}\rfloor}\ dx=4$$ and $$\int_{-\infty}^{\infty}e^{-|x|}\ dx=2$$

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Well I think $$\displaystyle \begin{align*} \int_{0}^{\frac{\pi}{2}} \theta \log^{3} (\tan\theta) \, d\theta &= \frac{7 \pi^{2}}{32} \zeta (3) + \frac{93}{16} \zeta (5). \end{align*}$$

is pretty interesting.

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