How do I solve integrals over $\mathrm d (\ln x)$? I am trying to solve integrals of the following form:
$$\int_{a}^{\infty}x^b\left[ e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right]d\ln x$$
Where $a,b$ are positive real numbers. My main aim is to make this into a form which is easier to numerically integrate, so I think that I need to change variables so that the integral is over a finite region (such as $x\rightarrow \frac{1}{x}$). However, I am not sure at all how to deal with the measure $\mathrm d(\ln x)$. If I change variables using something like $u= \ln x$, then the exponentials end up with exponentials in them again? Which does not help me simplify it at all.
How do I deal with this kind of integral? Should I attempt to remove the $d \ln x$ at all? I have struggled to come up with a suitable change of variables which makes the integral any easier to numerically/analytically solve.
Any suggestions or help would be greatly appreciated!
 A: You can use the fact which is associated with Stieltjes-Riemann integral i.e.
$$
d\ln x =(\ln x)'dx =\dfrac1x dx.
$$
Then we have
$\int_{a}^{\infty}x^b\left[ e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right]d\ln x=\int_{a}^{\infty}\dfrac1x x^b\left[ e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right]d x=
\int_{a}^{\infty} x^{b-1}\left[ e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right]d x=
\int_{a}^{\infty} x^{b-1} e^{\frac{x}{500}}-\int_{a}^{\infty} x^{b-1}e^{\frac{-x^2}{700}}dx.
$
The result you should obtain thanks to method of integration by parts. Can you end the calculations?
A: Since $\;d\ln x=\left(\ln x\right)^\prime dx=\frac{1}{x}dx\;,$ we get that
$\int_{a}^{+\infty}x^b\left(e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right)d\ln x=\int_{a}^{+\infty}x^{b-1}\left(e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right)dx\;.$
Since $\;x^{b-1}\left(e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right)>0\;$ for all $\;x>0\;$ and $\;\lim\limits_{x\to +\infty} x^{b-1}\left(e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right)=+\infty\;,$ it follows that
$\int_{a}^{+\infty}x^{b-1}\left(e^{\frac{x}{500}}-e^{\frac{-x^2}{700}}\right)dx=+\infty\;.$
We do not need to apply the method of integration by parts, indeed the integrand function is positive and divergent to infinity, so the integral is $+\infty\;.$
