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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1Could you please be more specific with what kind of integrals you're looking for? – teadawg1337 Dec 08 '14 at 21:09
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@teadawg1337 , integrals with unexpected results or integrals with just overall interesting results. – Kurtbusch Dec 08 '14 at 21:10
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1You should look for a course in complex analysis! – Raclette Dec 08 '14 at 21:10
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1See: the highly voted questions with the relevant tag(s). – Dec 08 '14 at 21:34
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2If you mean "other integrals where $\pi$ shows up for no apparent reason" - yes, definitely, that's part of $\pi$'s job. – Milo Brandt Dec 08 '14 at 22:58
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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.
Ron Gordon
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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.
Olivier Oloa
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Could you share more on how to evaluate the interesting family of integrals that you mentioned? – Lucian Dec 09 '14 at 22:08
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@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 09 '14 at 22:23
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Actually, the site-rules explicitly allow and even encourage users to ask-and-answer their own questions. – Lucian Dec 09 '14 at 22:28
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@Lucian It is done here:http://math.stackexchange.com/questions/1060107/a-family-of-fractional-part-integrals-with-the-digamma-function . Thank you. – Olivier Oloa Dec 10 '14 at 00:38
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${\tt\mbox{Sophomore's Dream}}$: Discovered by Johann Bernoulli in 1697 !!!.
$$ \int_{0}^{1}x^{-x}=\sum_{n\ =\ 1}^{\infty}n^{-n} $$
Felix Marin
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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.
FDP
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$$\begin{align} \int_0^\infty\exp\Big(-\sqrt[n]x\Big)dx&=n! \\ \\ \int_0^1\Big(1-\sqrt[n]x\Big)^mdx&={m+n\choose n}^{-1}={m+n\choose m}^{-1} \\ \\ \int_0^1\ln\Big(1-\sqrt[n]x\Big)dx&=-H_n \\ \\ \lim_{n\to\infty}n\Big(1-\sqrt[n]x\Big)&=-\ln x \end{align}$$
Lucian
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$$\int_{\mathbf{R}}\frac{dx}{1+x^2}=\pi$$ and $$\frac{22}{7} - \pi = \int_0^1 \frac{(x-x^2)^4}{1+x^2}dx.$$
Mister Benjamin Dover
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6Actually, $\int_{\mathbb R} \frac{e^{itx}}{1+x^2} dx = e^{-|t|}\pi$ for any $t\in \mathbb R$. Note that the integrand is complex but the answer is real.. – Raclette Dec 08 '14 at 21:15
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$$\int_0^{\infty}\left(\frac{3}{4}\right)^{\lfloor{x}\rfloor}\ dx=4$$ and $$\int_{-\infty}^{\infty}e^{-|x|}\ dx=2$$
homegrown
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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.
Joao
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