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Can anyone help me evaluate the following integral: $$ I(a)=\int_{a}^{\infty} e^{t}\, t^{-a}\, \Gamma(a-1,t)\, dt, $$ where $a\in(0,1)$ is a fixed parameter and $\Gamma(\cdot,\cdot)$ denotes the upper incomplete Gamma function?

One idea is to use the integral representation $$ e^{t} t^{-a}\Gamma(a-1,t)=\dfrac{1}{t}\int_0^{\infty}\dfrac{e^{-ty}}{(1+y)^{2-a}}dy, $$ which gives $$ I(a)=\int_0^\infty\dfrac{E_{1}(ay)}{(1+y)^{2-a}}dy, $$ where $E_1(z)$ denotes the basic exponential integral function. I am not sure how to proceed from here though.

Another idea is to use the integral representation $$ e^{t} t^{-a}\Gamma(a-1,t)=\dfrac{1}{t\Gamma(2-a)}\int_0^{\infty}\dfrac{y^{1-a} e^{-y}}{(y+t)}dy, $$ which gives $$ I(a)=\dfrac{1}{\Gamma(2-a)}\int_0^\infty y^{-a} e^{-y}\log(1+y/a) dy; $$ note that $a\in(0,1)$ so the integral is convergent.

Jason
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1 Answers1

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Using this series expansion and the Pocchhammer symbol $(u)_v$. We expand out the $k=0$ term of the sum to prevent $0$ division error:

$$\int e^t t^{-a}\left(\Gamma(a-1)-\frac{t^{a-1}}{(a-1)e^t}\left(1+\sum_{k=1}^\infty\frac{t^k}{(a)_k}\right)\right)dt=\int\Gamma(a-1)e^t t^{-a}dt-\int \frac{dt}{(a-1)t}-\int\sum_{k=1}^\infty\frac{t^{k-1}}{(a-1)(a)_k}dt$$

Now integrate and simplify the sum with the $\,_2\text F_2$ hypergeometric function to get:

$$(-1)^a\Gamma(a-1)\Gamma(a-1,-t)-\frac{\ln(t)}{a-1}-\frac{t\,_2\text F_2(1,1;2,a+1;t)}{(a-1)a}$$

This answer will be polished later, but the general idea is there.

Тyma Gaidash
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