$$\int_{0}^{+\infty}\dfrac{e^{-ax^p} - e^{-bx^p}}{x}dx$$

I would use the identity of Frullani given by:

$$ \int_{0}^{+\infty} \dfrac{f(ax) - f(bx)}{x} = \left[f (0) - f (+\infty)\right] \log \left(\frac{b}{a}\right) $$ with $ a, b > 0$

and the result is not the same if we use the technique to transform this integral into a double integral.


If you set $f(x) = e^{-x^p}$, then the integral is $\displaystyle\int_{0}^{\infty}\dfrac{f(a^{1/p}x)-f(b^{1/p}x)}{x}\,dx = \ln \dfrac{b^{1/p}}{a^{1/p}} = \dfrac{1}{p}\ln \dfrac{b}{a}$.


Here is the comparison with the double integral using $f(x)=e^{-x^2}$:

$$\int_{0}^{\infty}\frac{e^{-ax^p}-e^{-bx^p}}{x}\,dx = \int_{0}^{\infty}\frac{f(a^{1/p}x)-f(b^{1/p}x)}{x}\,dx = \int_{0}^{\infty} \left(\int_{b^{1/p}}^{a^{1/p}}f'(xy)\,dy\right)dx = \\\int_{0}^{\infty} \left(\int_{b^{1/p}}^{a^{1/p}}-p(xy)^{p-1}e^{-(xy)^{p}}\,dy\right)dx = -\int_{b^{1/p}}^{a^{1/p}} \left(\int_{0}^{\infty} p(xy)^{p-1} e^{-(xy)^{p} }\,dx\right)dy.$$

We can change the order of integration because the integrand is monotone.

Making the change of variables $u=xy$ in the inner integral we get

$$\int_{0}^{\infty}\frac{e^{-ax^p}-e^{-bx^p}}{x}\,dx=-\int_{b^{1/p}}^{a^{1/p}}\frac1{y} \left(\int_{0}^{\infty} pu^{p-1} e^{-u^{p} }\,dx\right)dy.$$

Changing variables again with $u^p=z$ we get

$$\int_{0}^{\infty}\frac{e^{-ax^p}-e^{-bx^p}}{x}\,dx=-\int_{b^{1/p}}^{a^{1/p}}\frac1{y} \left(\int_{0}^{\infty} e^{-z }\,dz\right)dy.=-\int_{b^{1/p}}^{a^{1/p}}\frac1{y} dy= \frac1{p}\ln{\frac{b}{a}}$$


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