Dose exist closed form for $ I=\int_{x=0}^{\alpha}x^{n-1}e^{-x\beta}\gamma(m,(\alpha-x)\zeta) $

Question

A am working with gamma random variable. While I am simplify the derived expression, I stuck in the following integral.

Let $n$ and $m$ be positive integers and $x, \alpha, \zeta$ $\beta$ real numbers.

I would like to simplify the following integral $I$

$$ I=\int_{x=0}^{\alpha}x^{n-1}e^{-x\beta}\gamma\Big(m,(\alpha-x)\zeta\Big)dx, $$

where $\gamma(n,x)$ is the incmplet lower gamma function define by

$$ \gamma(n,x)=\int_{t=0}^{x}t^{n-1}e^{-t}dt. $$

Thanks

Answer 1

We need to evaluate:

$$I=\int_0^{\alpha}x^{n-1}e^{-\beta x} \int_0^{(\alpha-x)\zeta}t^{m-1}e^{-t}dt dx$$

First, let's make the following substitution:

$$x=\alpha y$$

Then we have:

$$I=\alpha^n \int_0^1 y^{n-1}e^{-b y} \int_0^{c(1-y)}t^{m-1}e^{-t}dt dy$$

Where $b=\alpha \beta$, $c=\alpha \zeta $, which allows us to reduce the number of parameters.

Now let's try to do the same with the other variable.

$$t=c(1-y)u$$

$$I=\alpha^n c^m \int_0^1 y^{n-1} (1-y)^m e^{-b y} \int_0^1 u^{m-1}e^{-c(1-y)u}du dy$$

Maybe a series expansion of the second integral could work:

$$I=\alpha^n c^m \sum_{k=0}^\infty \frac{(-1)^k c^k}{k!} \int_0^1 y^{n-1} (1-y)^{m+k} e^{-b y} \int_0^1 u^{m+k-1}du dy$$

$$I=\alpha^n c^m \sum_{k=0}^\infty \frac{(-1)^k c^k}{k!(m+k)} \int_0^1 y^{n-1} (1-y)^{m+k} e^{-b y} dy$$

The last integral gives us:

$$\int_0^1 y^{n-1} (1-y)^{m+k} e^{-b y} dy=(n-1)! (m+k)! \, _1\tilde{F}_1(n;n+m+k+1;-b)$$

Where $\, _1\tilde{F}_1$ is a generalized regularized hypergeometric function.

In this way we obtain a (formal) series:

$$I=(n-1)! \alpha^n c^m \sum_{k=0}^\infty \frac{(-1)^k (m+k-1)! c^k}{k!} \, _1\tilde{F}_1(n;n+m+k+1;-b)$$

The question of convergence is a different animal, however I can certainly say that there are values of $b,c$ such that the series converges.

Here are a couple of examples in Mathematica (I use my third integral expression to compare with the series):

In[384]:= n = 6; 
m = 8; 
b = 4; 
c = 4; 
NIntegrate[y^(n - 1)*(1 - y)^m*Exp[(-b)*y]*u^(m - 1)*Exp[(-c)*(1 - y)*u], {y, 0, 1}, {u, 0, 1}, WorkingPrecision -> 10]
N[(n - 1)!*Sum[((-c)^k*(m + k - 1)!*Hypergeometric1F1Regularized[n, n + m + k + 1, -b])/k!, {k, 0, 100}], 10]
Out[388]= 1.719760455*10^-7
Out[389]= 1.719721615*10^-7

In[415]:= $MaxExtraPrecision =1000;
n = 3; 
m = 11; 
b = 1/5; 
c = 1/3; 
NIntegrate[y^(n - 1)*(1 - y)^m*Exp[(-b)*y]*u^(m - 1)*Exp[(-c)*(1 - y)*u], {y, 0, 1}, {u, 0, 1}, WorkingPrecision -> 10]
N[(n - 1)!*Sum[((-c)^k*(m + k - 1)!*Hypergeometric1F1Regularized[n, n + m + k + 1, -b])/k!, {k, 0, 100}], 10]
Out[420]= 0.00006265635078
Out[421]= 0.00006265694467

In conclusion, this series is obviously complicated, and it's better to evaluate the integral numerically. I'm not sure a closed form exists at all, though of course, we can try other ways to derive it.