# Definite integrals with interesting results [closed]

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?

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

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• Could you please be more specific with what kind of integrals you're looking for? – teadawg1337 Dec 8 '14 at 21:09
• @teadawg1337 , integrals with unexpected results or integrals with just overall interesting results. – Kurtbusch Dec 8 '14 at 21:10
• You should look for a course in complex analysis! – Raclette Dec 8 '14 at 21:10
• See: the highly voted questions with the relevant tag(s). – user147263 Dec 8 '14 at 21:34
• If you mean "other integrals where $\pi$ shows up for no apparent reason" - yes, definitely, that's part of $\pi$'s job. – Milo Brandt Dec 8 '14 at 22:58

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

• 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.. – Raclette Dec 8 '14 at 21:15

One I just evaluated

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

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

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

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.

• Okay. This family what you've created. Not bad. +1. – user153012 Dec 9 '14 at 1:30
• @user153012 Thank you. – Olivier Oloa Dec 9 '14 at 20:08
• Could you share more on how to evaluate the interesting family of integrals that you mentioned? – Lucian Dec 9 '14 at 22:08
• @Lucian Yes. Maybe, you could ask a related question so I would have more room than the one for this comment? Thanks. – Olivier Oloa Dec 9 '14 at 22:23
• Actually, the site-rules explicitly allow and even encourage users to ask-and-answer their own questions. – Lucian Dec 9 '14 at 22:28

${\tt\mbox{Sophomore's Dream}}$: Discovered by Johann Bernoulli in 1697 !!!.

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

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

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

Math.stackexchange is full of such beautiful formulae.

• but there seems to be room for more... – Jaume Oliver Lafont Feb 15 '16 at 11:17

\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}

$$\int_0^{\infty}\left(\frac{3}{4}\right)^{\lfloor{x}\rfloor}\ dx=4$$ and $$\int_{-\infty}^{\infty}e^{-|x|}\ dx=2$$

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.