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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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closed as too broad by mrf, user147263, AlexR, Gyu Eun Lee, Thomas Dec 9 '14 at 0:39

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Could you please be more specific with what kind of integrals you're looking for? $\endgroup$ – teadawg1337 Dec 8 '14 at 21:09
  • $\begingroup$ @teadawg1337 , integrals with unexpected results or integrals with just overall interesting results. $\endgroup$ – Kurtbusch Dec 8 '14 at 21:10
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    $\begingroup$ You should look for a course in complex analysis! $\endgroup$ – Raclette Dec 8 '14 at 21:10
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    $\begingroup$ See: the highly voted questions with the relevant tag(s). $\endgroup$ – user147263 Dec 8 '14 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 '14 at 22:58
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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 '14 at 21:15
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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 '14 at 1:30
  • $\begingroup$ @user153012 Thank you. $\endgroup$ – Olivier Oloa Dec 9 '14 at 20:08
  • $\begingroup$ Could you share more on how to evaluate the interesting family of integrals that you mentioned? $\endgroup$ – Lucian Dec 9 '14 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 '14 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 '14 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 '16 at 11:17
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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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$$\int_0^{\infty}\left(\frac{3}{4}\right)^{\lfloor{x}\rfloor}\ dx=4$$ 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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