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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

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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.

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  • $\begingroup$ 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
    Commented Oct 30, 2020 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
    Commented Oct 30, 2020 at 21:45
  • $\begingroup$ I corrected this $\endgroup$
    – am301
    Commented Nov 5, 2020 at 6:29
  • $\begingroup$ Do you know what the units would be? $\endgroup$
    – N A
    Commented Dec 6, 2020 at 3:08
  • $\begingroup$ This question is not clear $\endgroup$
    – am301
    Commented Dec 7, 2020 at 4:53

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