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This question is an offspring of the question A closed form for: $\int_{0}^{\infty} \frac{1}{(x-\log x)^2}dx$

In amongst the comments and incomplete answers to this question, a conjecture was raised on the equivalence between two different generalized definite integrals for real $p \ge 2$, that is $$\int_0^{\infty } \frac{1}{(x-\log (x))^p} \, dx= \frac{ 1}{\Gamma(p)} {\int_0^{\infty } \frac{\Gamma(x+1)}{x^{x-p+2}} \, dx}\tag{1}$$

In an attempt to prove this I have noted the known result

$$\int_0^\infty t^b e^{-a t} \,dt=\frac{\Gamma(b+1)}{a^{b+1}}\tag{2}$$

proved by integration by parts multiple times in the case of integer $b$ (see for example https://en.wikipedia.org/wiki/Gamma_function#Applications)

By utilizing (2) and changing the order of integration on the r.h.s. of (1) I was led to the following conjecture for $t>0$ and $s>0$

$$\frac{1}{\Gamma(s)}\int_0^\infty x^{s-1} t^{x} e^{-x t} \,dx=\frac{1}{(t-\log(t))^s}$$

Mathematica 12.0 manages to calculate this integral quite happily, but I'm none the wiser how it solves this, even in the case of integer values of $s$. Any ideas or relevant references for solution of integrals of this kind?

James Arathoon
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1 Answers1

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$(2)$ is proven simply by substitution $t=x/a$ and the definition of $\Gamma(b+1)$.

Your last formula is an application of this, as $t^x e^{-xt}=e^{-(t-\log(t))x}$.

metamorphy
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  • @James: I would like to proof, following the way of Sasha using the Plancherel theorem. This should work, but I have not the time. I suggest you to try this way. – stocha Sep 23 '19 at 07:56