0
$\begingroup$

I've come up with the following integral as part of an equation for a generalized flow model for groundwater flow. Unfortunately it doesn't converge. Does anyone know of a way I can manipulate it in order to make it solvable?

$$ \int_0^t \frac{1}{x^{w/2}} e^{-c/x} dx$$

$1 \leq w \le 3$ and $c$ is a positive number

$\endgroup$
5
$\begingroup$

We make the transformation: $y=\frac{c}{ x}$. We have: $dx=-\frac{cdy}{ y^2}$, and we can put the integral in the form: $$ I=c^{1-\frac{w}{2}}\int_{\frac{c}{t}}^\infty y^{\frac{w}{2}-2}e^{-y}dy=c^{1-\frac{w}{2}}\Gamma\bigg(\frac{w}{2}-1,\frac{c}{t}\bigg)$$ where $\Gamma$ is the incomplete gamma function.

$\endgroup$
10
  • $\begingroup$ Thanks for the help. The transformation makes sense easily but could you show me how the solution for the integral is derived? Will the incomplete gamma function be defined when (w/2 - 1) is less than 0? $\endgroup$ – N A Oct 30 '20 at 21:21
  • $\begingroup$ I think I see now – by definition $$ \int_{x}^\infty y^{a-1}e^{-y}dy= \Gamma\bigg(a,x\bigg) $$ The $ c^{3-\frac{w}{2}} $ I’m not sure about though. $\endgroup$ – N A Oct 30 '20 at 21:45
  • $\begingroup$ I corrected this $\endgroup$ – am301 Nov 5 '20 at 6:29
  • $\begingroup$ Do you know what the units would be? $\endgroup$ – N A Dec 6 '20 at 3:08
  • $\begingroup$ This question is not clear $\endgroup$ – am301 Dec 7 '20 at 4:53

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.