14

I know that both $$\int_{-\infty}^\infty \frac{\cos(x)}{x^2+1} \, \text{dx} = \frac{\pi}{e}$$ and $$\int_{-\infty}^\infty \frac{x\sin(x)}{x^2+1} = \frac{\pi}{e}$$ But what about $\frac{e}{\pi}$? Is there a non-trivial definite integral which evaluates to that, but still where the integrand is composed of elementary functions like the two above? The only constraint on the integrand is that it is an elementary function. It does not need to have an elementary antiderivative. Also, the value of the integral should not be obvious at first glance. The integrals mentioned above are certainly not, which makes the answer so beautiful in my opinion.

By trivial I mean integrals like $\int_{-\infty}^1 \frac{e^x}{\pi} \, \text{d}x = \frac{e}{\pi}$, which quite obviously converges to the desired value.

Edit: I have been asked to clarify my question as it is too broad. The two integrals mentioned above are, in my opinion, beautiful because they both converges to a fraction containing two such fundamental constants, even though it is not obvious at all that they should at first glance. Then, out of curiosity, I wondered if there are similar integrals which converges to $\frac{e}{\pi}$. The answer @aleden has written is certainly interesting, but some of the magic disappears when both $\pi$ and $e$ are in the integrand. I have added some more information in the paragraphs above. I hope this cleared my question a bit, but it is broad by its nature.

Arctic Char
  • 16,007
Markus
  • 337
  • 3
    Getting $\pi$ in the numerator is easy (residue theorem), getting $\pi$ in the denominator is harder. I don't see how 7 people can think this is a good question, it basically just says "show me an integral that has this rather arbitrary value". Why $\frac{e}{\pi}$? – Winther Oct 16 '18 at 23:42
  • 4
    Isn't it such beautiful value? Imo this question is interesting because as you also said it is harder to get $\pi$ in the denominator. – Zacky Oct 16 '18 at 23:45
  • 3
    Yes but it's kind of weird to ask for a function that lands on what is, an arbitrary value. It's like asking for a nontrivial polynomial that has a root at 1729, because it's a taxicab number and thus cool. Of course, polynomials are trivial, but what defines "trivial" here? At least, that's what Winther seems to be getting at. – Nicholas Pipitone Oct 16 '18 at 23:49
  • 2
    This type of question appeared here before, and looks like it was well received: https://math.stackexchange.com/questions/168485/help-find-hard-integrals-that-evaluate-to-59 – Zacky Oct 17 '18 at 00:01
  • @Dahaka Question was also closed for being too broad... – Frank W Oct 17 '18 at 01:23
  • I think the question could be improved by giving explicit, precise and restrictive constraints on what functions are allowed as integrands. E.g. can a rational expression in $x$ and $\sin(x)$ with only rational coefficients work? – Torsten Schoeneberg Oct 17 '18 at 05:04
  • 1
    @Dahaka That one is truly beautiful. Thank you for sharing it. Where did you find that integral or how did you derive it? If you have more (non-trivial) integrals with $\pi$ in the denominator, I would be very glad if you shared them as well. – Markus Oct 18 '18 at 21:27

1 Answers1

2

By Frullani's Theorem:

$$\frac{1}{f(0)-f(\infty)}\int_0^\infty \frac{f(e^{\pi} x)-f(e^e x)}{x}dx=\frac{e}{\pi}$$ for appropriate functions $f(x)$.

EDIT: Alternatively, you can look for a function $f(x)$ that satisfies $\lim_{x\to\infty}f(x)=\frac{e}{\pi}$ such as $f(x)=\frac{ex}{\pi x+1}$ and use:

$$\frac{1}{\ln(\frac{a}{b})}\int_0^\infty \frac{eax(\pi bx+1)-ebx(\pi ax+1)}{x(\pi ax+1)(\pi bx+1)}dx=\frac{e}{\pi}$$

aleden
  • 4,007
  • 13
    So, I think it is not so surprising we can get $e/\pi$ when your integrand has $e$ in the numerator and $\pi$ in the denominator. – GEdgar Oct 17 '18 at 11:56