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How do I prove this

$$\int_{0}^{\infty}{e^{-x^n}-e^{-x^m}\over x\ln{x}}dx=\color{blue}{\ln{\left(m\over n\right)}}.\tag1$$

I know of the standard integral

$$\int_{0}^{1}{x^m-x^n\over \ln{x}}dx=\ln\left({m+1\over n+1}\right)\tag2$$

I can't seem to find a suitable substitution for $(1)$

Can someone give a hint please? Thank you.

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Take $\log\left(x\right)=v$. We have \begin{align} I&=\int_{0}^{\infty}\frac{\exp\left(-x^{n}\right)-\exp\left(-x^{m}\right)}{x\log\left(x\right)}\ dx\\[10pt] &=\int_{-\infty}^{\infty}\frac{\exp\left(-e^{vn}\right)-\exp\left(-e^{vm}\right)}{v}\ dv\\[10pt] &=\int_{0}^{\infty}\frac{\exp\left(-e^{vn}\right)-\exp\left(-e^{vm}\right)}{v}\ dv-\int_{0}^{\infty}\frac{\exp\left(-e^{-vn}\right)-\exp\left(-e^{-vm}\right)}{v}dv \end{align} so if we apply the Frullani's theorem to the function $f\left(x\right)=\exp\left(-e^{x}\right)$ and $g\left(x\right)=\exp\left(-e^{-x}\right)$ respectively we get

$$I=\frac{1}{e}\log\left(\frac{m}{n}\right)-\left(\frac{1}{e}-1\right)\log\left(\frac{m}{n}\right)=\color{red}{\log\left(\frac{m}{n}\right)}$$

as wanted.

Addendum. It is interesting to note that we can easily generalize the result. We have the following:

Theorem. If $f:\left(0,\infty\right)\rightarrow\mathbb{R} $ is a function such that $\lim_{x\rightarrow0}f\left(x\right)=f\left(0\right)\in\mathbb{R} $ and $\lim_{x\rightarrow\infty}f\left(x\right)=f\left(\infty\right)\in\mathbb{R} $ and is integrable over any interval $0<A\leq x\leq B<\infty $, then for all $m,n>0 $ we get $$\int_{0}^{\infty}\frac{f\left(x^{n}\right)-f\left(x^{m}\right)}{x\log\left(x\right)}dx=\left(f\left(0\right)-f\left(\infty\right)\right)\log\left(\frac{m}{n}\right). $$

Proof: We have $$I=\int_{0}^{\infty}\frac{f\left(x^{n}\right)-f\left(x^{m}\right)}{x\log\left(x\right)}dx\overset{\log\left(x\right)=v}{=}\int_{-\infty}^{\infty}\frac{f\left(e^{vn}\right)-f\left(e^{vm}\right)}{v}dx $$ $$=\int_{0}^{\infty}\frac{f\left(e^{vn}\right)-f\left(e^{vm}\right)}{v}dx-\int_{0}^{\infty}\frac{f\left(e^{-vn}\right)-f\left(e^{-vm}\right)}{v}dx $$ and now since we have the hypothesis of the classic Frullani's theorem we get $$\begin{align} I= & \left(f\left(1\right)-f\left(\infty\right)\right)\log\left(\frac{m}{n}\right)-\left(f\left(1\right)-f\left(0\right)\right)\log\left(\frac{m}{n}\right)\\ = & \left(f\left(0\right)-f\left(\infty\right)\right)\log\left(\frac{m}{n}\right).\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square \end{align}$$

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    $\begingroup$ So Nice... +(1) $\endgroup$ – Behrouz Maleki Jul 1 '16 at 14:04
  • $\begingroup$ Any continuous function $(0,\infty) \to \mathbb{R}$ is integrable over any interval $[A,B]$ with $0 < A < B < \infty$, mainly because $[A,B]$ is compact. $\endgroup$ – Najib Idrissi Jul 1 '16 at 14:15
  • $\begingroup$ @NajibIdrissi Indeed continuity is a unnecessary assumption. I change the statement of the theorem. Thank you. $\endgroup$ – Marco Cantarini Jul 1 '16 at 17:25
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Hint. Assume we have $n>0,\, m>0$ and set $$ f(n,m):=\int_{0}^{\infty}{e^{-x^n}-e^{-x^m}\over x\ln{x}}dx. $$ One may just derivate with respect to $n$: $$ \frac{\partial}{\partial n}f(n)=-\int_{0}^{\infty}x^{n-1}e^{-x^n}dx=-\frac1n $$ then integrating with respect to $n$, using $f(m,m)=0$, gives

$$ \int_{0}^{\infty}{e^{-x^n}-e^{-x^m}\over x\ln{x}}dx={\ln{\left(m\over n\right)}} $$

as announced.

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    $\begingroup$ This is too concise for me @Olivier Oloa, I don't get it. How did you arrive at the answer so quick? $\endgroup$ – gymbvghjkgkjkhgfkl Jul 1 '16 at 1:46
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    $\begingroup$ @Chinacat We are allowed to differentiate under the integral sign since all conditions are fulfilled for it, just take $e^{-x}$ as the dominated function of the partial derivative of the integrand over $[1,\infty)$. Have you already checked by your own that $\frac{\partial}{\partial n}{e^{-x^n}-e^{-x^m}\over x\ln{x}}=-x^{n-1}e^{-x^n}$, if not please do it, it is an important point. Then, with $u=x^n$, $du=nx^{n-1}dx$ and you get $\int_{0}^{\infty}x^{n-1}e^{-x^n}dx=\frac1n \cdot \int_{0}^{\infty}e^{-u}du=\frac1n$. $\endgroup$ – Olivier Oloa Jul 1 '16 at 4:51
  • $\begingroup$ I don't understand the last part, integrating with respect to n using f(m,m) and how this leads to the answer. Is the integration just used for reversing the derivation? Does Oliver then do the same with m as n, however getting m in the nominator. His approach shows how to get the n into the denominator. I am more interested in the idea, how it works than the verification, I would verify it myself. $\endgroup$ – Imago Jul 1 '16 at 15:02
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    $\begingroup$ Since $f'(n)=-1/n$ over $n>0$ then $f(n)=-\ln n+C$, where $C$ is a constant with respect to $n$, but a priori $C$ is not a constant with respect to $m$, that is $C:=C(m)$, giving $f(n)=-\ln n +C(m)$. Now $f(m)=0$, thus $0=-\ln m +C(m)$, $C(m)=\ln m$, thus $f(n)=-\ln n +\ln m=\ln (m/n)$ as wanted. Tell me if it is Ok now. Thanks. $\endgroup$ – Olivier Oloa Jul 1 '16 at 18:14
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    $\begingroup$ Oh, this is even more brilliant that I thought. Wow :), I never have used switching the order (integration / differentiation) and computing C, so it fits the functions properties together before, this offers whole new ways for me looking at integrals :) $\endgroup$ – Imago Jul 3 '16 at 18:38

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