How do I prove this $\int_{0}^{\infty}{e^{-x^n}-e^{-x^m}\over x\ln{x}}dx={\ln{\left(m\over n\right)}}?$

Question

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.

Answer 1

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 $$\begin{align} I= & \left(f\left(1\right)-f\left(\infty\right)\right)\log\left(\frac{m}{n}\right)-\left(f\left(1\right)-f\left(0\right)\right)\log\left(\frac{m}{n}\right)\\ = & \left(f\left(0\right)-f\left(\infty\right)\right)\log\left(\frac{m}{n}\right).\\ & \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square \end{align}$$

Answer 2

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.