# How can we prove that this integral converges?

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?   • Hint: look at the integrand for $t\to+\infty$. For $x>0$ the integrand goes to $\frac{1}{r}$ for $t\to+\infty$, hence the integral diverges. For $x\lt0$ the integrand behaves as $\frac{1}{a} e^{-|x| t}$ and the integral is convergent. – Dr. Wolfgang Hintze Apr 18 at 12:51
• In the difference of the two integrands the constant terms cancel, and you are left with integrable terms. – Dr. Wolfgang Hintze Apr 18 at 19:59
• Replace the difference of the two intergrals by one intergral over the difference of the integrands. Start with intergrals with a finite upper limit A, take the difference of these, transfer the difference to the integrand and only in the end take the limit $A\to \infty$. – Dr. Wolfgang Hintze Apr 19 at 4:16

## 1 Answer

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

$$f = \frac{1}{y} \frac{\sqrt{r^2+1} \log \left(\frac{\sqrt{a^2+1}+1}{a}\right)+\log \left(\frac{\sqrt{\left(a^2+1\right) \left(r^2+1\right)}+a r-1}{a+r}\right)-\sinh ^{-1}(r)}{r \sqrt{r^2+1}}$$

• Ok, but I'm interested only in the case of $x>0$ and difference between such two integrals as shown on Picture 1. Also, about first part of your proof: how can we estimate convergence through integrand? Shouldn't we investigate limits for antiderivative instead? – Tag Apr 18 at 13:39
• Consider the much simpler example: $i_1=\int_0^\infty \frac{ 1}{1+x}$ and $i_2=\int_0^\infty \frac{ 1}{2+x}$. Each integral is divergent but the difference can be given a meaning by integrating the difference of the integrands which converges. – Dr. Wolfgang Hintze Apr 19 at 4:24
• Still can't understand how to show this more rigorous with proper math. – Tag Apr 19 at 20:02
• Sorry, but I have descibed here in several occasions how you can transform your question into proper math. Your $\infty - \infty$ surely is not "proper math". – Dr. Wolfgang Hintze Apr 20 at 7:02