I cannot find a closed form solution for the following integral: $$\int_0^t\exp\left[- \frac{a(\delta+1)(t-x)}{(\delta-(t-x))}\right] \, b\exp[-bx] dx$$ So, to evaluate this integral, I have used numerical integration (in Matlab). Note that $a$, $b$, $\delta$ and $t$ are all positive. When I set $\delta<t$ the result is infinity. However, if $t=\delta$ the result is not infinity; ex: $a=0.1$, $b=1$, and $t=4$.

I suspect that the problem results from the denominator $\delta-(t-x)$, which is equal to $0$ if $x= t-\delta$.

My question: does the above mean that the integral diverges and goes to infinity for $\delta <t$ ? or it is the numerical approach used that is not good in this case? if so, is there any other approach to compute the integral ?

  • $\begingroup$ what is the meaning of this brackets $$[...]$$? $\endgroup$ – Dr. Sonnhard Graubner Feb 21 '17 at 9:52
  • $\begingroup$ $\exp[..]$ is equivalent to $\exp(..)$. Maybe it is a bad choice to use this notation. Please feel free to edit the question. $\endgroup$ – din Feb 21 '17 at 9:54
  • $\begingroup$ can you say something about the parameters? $\endgroup$ – Dr. Sonnhard Graubner Feb 21 '17 at 10:02
  • $\begingroup$ This is not an integral unless you decide something for $d$, say $dx$? $\endgroup$ – mathreadler Feb 21 '17 at 10:11
  • $\begingroup$ Thank you for pointing out this issue. I have edited the question. $\endgroup$ – din Feb 21 '17 at 10:15

The first thing to do is to change variables $t-x=tz$, yielding $$ I=\int_0^t\exp\left[- \frac{a(\delta+1)(t-x)}{(\delta-(t-x))}\right] \, b\exp[-bx]dx=bt e^{-bt}\int_0^1 dz \exp\left[- \frac{a(\delta+1)tz}{\delta-tz}\right] e^{btz} $$ $$ \propto \int_0^1 dz \exp\left[- \frac{Az}{B-z}\right] e^{Cz}\ , $$ where $A$, $B$ and $C$ are the only combinations of parameters that are really relevant. Now, if $0<B<1$, the integrand has a one-sided divergence as $z\to B$ (essential singularity) inside the integration range which is non-integrable. As $B=\delta/t$, you have your answer. If $B$ is exactly equal to one, the integral is perfectly fine as $$ \lim_{z\to 1^-} \exp\left[-\frac{Az}{1-z}\right]=0\ , $$ so there are no divergences inside the integration range.

  • $\begingroup$ Just to make sure I understand your answer, for $\delta < t$ having $I$ equals to infinity is normal, meaning that the integral diverges if $\delta < t$, ? $\endgroup$ – din Feb 21 '17 at 11:59
  • $\begingroup$ Yes, that's correct $\endgroup$ – Pierpaolo Vivo Feb 21 '17 at 12:07

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