0
$\begingroup$

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?

Picture 1

Picture 2

Picture 3

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dr. Wolfgang Hintze Apr 18 at 12:51
  • $\begingroup$ In the difference of the two integrands the constant terms cancel, and you are left with integrable terms. $\endgroup$ – Dr. Wolfgang Hintze Apr 18 at 19:59
  • $\begingroup$ 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 $A\to \infty$. $\endgroup$ – Dr. Wolfgang Hintze Apr 19 at 4:16
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Tag Apr 18 at 13:39
  • $\begingroup$ 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. $\endgroup$ – Dr. Wolfgang Hintze Apr 19 at 4:24
  • $\begingroup$ Still can't understand how to show this more rigorous with proper math. $\endgroup$ – Tag Apr 19 at 20:02
  • $\begingroup$ 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". $\endgroup$ – Dr. Wolfgang Hintze Apr 20 at 7:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.