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I’m looking for pairs of difficult definite integrals that are linked algebraically on a certain field without known change of variable or integration by parts from one integral to the other.

Here a ‘difficult’ integral means an integral without a known closed form.

Let’s see some examples. The second one is from Ramanujan.

$$ \int_0^1 \frac{\psi(x+1)+\gamma}{x} \:{\mathrm{d}}x = \int_0^1 \frac{x \ln x}{(1-x) \ln (1-x)} \:{\mathrm{d}}x \tag1$$

where $\psi:= \Gamma'/\Gamma$ is the digamma function and $\gamma$ denotes the Euler constant.

$$ \int_0^\infty\frac{t^{x}}{\Gamma(x+1)}{\mathrm{d}}x=e^{t}-\int_0^\infty\frac{e^{-tx}}{x(\pi^2 +\ln^2 x)}{\mathrm{d}}x, \quad t>0. \tag2 $$

Could you give other examples?

Thanks.

Addendum.

Usually, when you try to evaluate an integral in closed form, you make use of transformations to bring it to a new integral which is a well-known integral. You end up with an algebraic relation between your integrals. Let’s say your integrals are ‘chromatic’ with respect to a field. When two integrals are ‘chromatic’, evaluating one gives the answer for the other. Your current tools among your transformations are: change of variable and integration by parts. What if we suppress these common tools? What is left then?

This post is supposed to obtain some explicit examples of ‘chromatic’ difficult integrals that, as far as we know, are not linked from a change of variable or an integration by parts.

Olivier Oloa
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2 Answers2

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What about these pairs?

\begin{align} \nonumber(\alpha-\sin \alpha)&\int_{-\infty}^\infty\frac{d x}{(e^x-x-\cos \alpha)^2+(\alpha-\sin \alpha)^2}+\\ &(\pi-\alpha+\sin \alpha)\int_{-\infty}^\infty\frac{d x}{(e^x+x+\cos \alpha)^2+(\pi-\alpha+\sin \alpha)^2}=\pi,\quad 0<\alpha<\pi \end{align}

\begin{align} \nonumber(2\pi-\alpha+\sin \alpha)&\int_{-\infty}^\infty\frac{d x}{(e^x-x-\cos \alpha)^2+(2\pi-\alpha+\sin \alpha)^2}=\\ &(\pi-\alpha+\sin \alpha)\int_{-\infty}^\infty\frac{d x}{(e^x+x+\cos \alpha)^2+(\pi-\alpha+\sin \alpha)^2},\quad 0<\alpha<\pi \end{align} It seems they are not linked by a change of a variable or integration by parts, and as far as I know none of these integrals have a closed form.

  • Interesting! (+1) – Olivier Oloa Oct 15 '15 at 15:19
  • I could imagine that this integrals may be calculated along the same lines as this one here: http://math.stackexchange.com/questions/1055468/integral-int-infty-infty-fracdxexx12-pi2 or are there reason against that? – tired Oct 22 '15 at 14:37
  • @tired, actually that proof you mentioned is wrong, because it takes into account just a single pole and leaves behind infinitely many other poles as if they just did not exist. But of course the answer has scored 10 upvotes and 100 points bounty, so who cares? –  Oct 22 '15 at 16:13
  • u can make this proof rigouros, @robjohn wrote down an answer somewhere where he showed that all residues but once cancel... – tired Oct 22 '15 at 16:14
  • http://math.stackexchange.com/questions/45745/interesting-integral-related-to-the-omega-constant-lambert-w-function?lq=1 – tired Oct 22 '15 at 16:15
  • @tired, in this case one needs to consider the sum of 2 integrals for the poles to cancel each other. The easiest way to see this is to note that if all unwanted poles cancelled for the first integral alone, then in general one would get an answer with a nonzero imaginary part. –  Oct 22 '15 at 16:44
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$$ \int_0^{\infty} \frac{ dx }{ \Gamma (x) } = e + \int_0^{\infty} \frac{e^{-x}}{\pi^2 + \ln^2 x} \,dx = e + \frac{1}{\pi} \int_{-\pi/2}^{\pi/2} e^{\tan x} e^{ -e^{\ \tan x} } \,dx $$ According to Wikipedia, it is unknown whether this constant can be expressed in closed form.

user111187
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