Need to evaluate the integral $$ \int_a^b e^{1/x}\,\Gamma(m,1/x)\,dx $$ or equivalently $$ \int_{1/a}^{1/b} y^{-2}\,e^{y}\,\Gamma(m,y)\,dy, $$ where $m$ is an integer, and $0<a<b<\infty$. The integral is easily computable for $m=0$ and $m=1$. Not sure how to deal with negative $m$'s. Mathematica does compute it numerically (for specific $a$ and $b$). Is there a way to obtain the actual formula?

  • $\begingroup$ I think integration by parts should work, using the definition of the derivative of the incomplete gamma function wrt y given on the wikipedia page. $\endgroup$
    – Bitrex
    Apr 4, 2014 at 17:49
  • $\begingroup$ I don't see it, alas. $\endgroup$
    – Jason
    Apr 4, 2014 at 21:06

1 Answer 1


First, we use the definition of the incomplete gamma and transform it a little:

$$\Gamma (m,y)=\int_y^\infty p^{m-1} e^{-p}dp=y^m \int_1^\infty t^{m-1} e^{-y t} dt$$

Let's denote:


Taking the second form of your integral and changing the order of integration:

$$\int_{\alpha}^{\beta} y^{-2}\,e^{y}\,\Gamma(m,y)\,dy=\int_1^\infty t^{m-1} \int_{\alpha}^{\beta} y^{m-2}\,e^{-(t-1)y} ~dy~ dt$$

Now it actually depends on the values of $a, b, m$. If they are general real numbers, we get an incomplete Gamma again:

$$\int_{\alpha}^{\beta} y^{-2}\,e^{y}\,\Gamma(m,y)\,dy=\int_1^\infty \frac{t^{m-1}}{(t-1)^{m-1}} \left( \Gamma [ m-1,\alpha(t-1) ]- \Gamma [ m-1,\beta(t-1) ] \right) dt$$


$$\int_{\alpha}^{\beta} y^{-2}\,e^{y}\,\Gamma(n+1,y)\,dy=\int_0^\infty \left( 1+\frac{1}{z} \right)^{n} \left( \Gamma ( n,\alpha z )- \Gamma ( n,\beta z ) \right) dz$$

Mathematica shows a general closed form for each integral separately:

$$\int_0^\infty \left( 1+\frac{1}{z} \right)^n \Gamma ( n,\alpha z ) dz=-\frac{1}{ n \Gamma (-n)} \left\{ \alpha^{n} \Gamma (-n-1) \, _2F_2(1,n;n+1,n+2;\alpha)+ \\ + \frac{\pi n}{ \sin (\pi n)} \left(-\frac{ \pi }{\tan(\pi n)}+\frac{e^\alpha}{ \alpha}+ \log (-\alpha)- \log (\alpha)+ \Gamma (0,-\alpha)\right) \right\}$$

With the conditions:

$$ (\Re(\alpha)=0\land \Re(n)<0)\lor (\Re(\alpha)>0\land \Re(n)<1)$$

I think we can use Frullani's theorem here somehow:

$$\int_0^\infty \frac{f(\alpha z)-f(\beta z)}{z}dz=(f(0)-f(\infty) \log \left( \frac{\beta}{\alpha} \right)$$

For example, in the case $n=1$ the integral is:

$$\int_0^\infty \left( 1+\frac{1}{z} \right) \left( \Gamma ( 1,\alpha z )- \Gamma ( 1,\beta z ) \right) dz=\frac{1}{\alpha}-\frac{1}{\beta}+\log \left( \frac{\beta}{\alpha} \right) \\ \Re(\beta)>0 \land \Re(\alpha)>0$$

Edit: Unlike the form I obtained, Mathematica can't solve the original integral in general.


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