If we have integral in the form $\int_0^\infty \frac{1}{(r+a \exp (-xt) )]\sqrt{(1+(a \exp (-xt))^2)}} \ dt $ and if we take difference of such two integrals with the same $r>0$ and different (or the same) $a>0$ and $x>0$ in this two integrals, this difference will converge. See example on Picture 1. Can we prove that?
I really can't understand why integral on Picture 1 does converge but integral on Picture 2 doesn't. Looking at the antiderivative on Picture 3 no idea why is that happens. Could you help please?
We derive the exact expression for the integral.
The integral in question is
$$f = \int_0^{\infty } \frac{1}{\sqrt{(a \exp (-x t))^2+1} (a \exp (-x y)+r)} \, dt$$
First let us have a look at the convergence of the integral. For $x\gt 0$ the integrand goes to $\frac{1}{r}$ for $t\to+\infty$, hence the integral diverges. For $x\lt0$ the integrand behaves as $(\frac{1}{a} e^{-|x| t})^2 $ and the integral is convergent.
Hence we assume $x\lt0$. Letting $y=-x>0$ and substituting $t\to \frac{1}{y} \log(u)$, $dt \to \frac{du}{u y}$, $t\in(0,\infty) \to u\in(1,\infty)$ the integral becomes
$$f=\frac{1}{y} \int_1^\infty \frac{1}{ u \sqrt{a^2 u^2+1} (a u+r)}\,du$$
Mathematica finds
$$f = \frac{1}{y} \frac{\sqrt{r^2+1} \log \left(\frac{\sqrt{a^2+1}+1}{a}\right)+\log \left(\frac{\sqrt{\left(a^2+1\right) \left(r^2+1\right)}+a r-1}{a+r}\right)-\sinh ^{-1}(r)}{r \sqrt{r^2+1}}$$