# Integrating lower incomplete gamma function raised to the power $k$

I'm trying to solve the following integral:

$$\int_0^\infty \gamma(t,x)^k x^t e^{-x} \mathrm{d} x$$

I'm fighting with it for quiet a while and didn't get any result. Though, I do have the following:

Noting that $x^t e^{-x} = \mathrm{d}\gamma(t+1,x)/\mathrm{d}x$, and using the identity $\gamma(t,x) = \frac{1}{t}(\gamma(t+1,x) + x^t e^{-x})$, the integral can be written as \begin{eqnarray*} \int_0^\infty \gamma(t,x)^k x^t e^{-x} \mathrm{d} x &=& \int_0^\infty \gamma(t,x)^k \mathrm{d} \gamma(t+1,x) \\ &=&\frac{1}{t^k} \int_0^\infty (\gamma(t+1,x) + \mathrm{d} \gamma(t+1,x))^k \mathrm{d} \gamma(t+1,x) \end{eqnarray*}

Second try:

$$\int_0^\infty \gamma(t,x)^k x^t e^{-x} \mathrm{d} x = \int_0^\infty \int_0^x \int_0^x \cdots \int_0^x \prod_{i=1}^k y_i^{t-1} e^{y_i} x^t e^{-x} \mathrm{d} y_1 \mathrm{d} y_2 \cdots \mathrm{d} y_k \mathrm{d} x$$ Now, substitute $\{y_i = x \cdot z_i , \mathrm{d}y_i = x\cdot \mathrm{z_i}\}$, we have: \begin{align} &\int_0^\infty \int_0^1 \int_0^1 \cdots \int_0^1 \left(\prod_{i=1}^k z_i\right)^{t-1} x^{t(k+1)} e ^{-x(\sum_{i=1}^k z_i +1)}\mathrm{d} z_1 \mathrm{d} z_2 \cdots \mathrm{d} z_k \mathrm{d} x &\\ &= \int_0^\infty \int_0^1 \int_0^1 \cdots \int_0^1 \frac{\left(\prod_{i=1}^k z_i\right)^{t-1}}{\left(\sum_{i=1}^k z_i +1\right)^{t(k+1)}} \left[x\left(\sum_{i=1}^k z_i +1\right)\right]^{t(k+1)} e ^{-x(\sum_{i=1}^k z_i +1)} \mathrm{d} z_1 \mathrm{d} z_2 \cdots \mathrm{d} z_k \mathrm{d} x &\\ &= \Gamma(t(k+1)+1) \int_0^1 \int_0^1 \cdots \int_0^1 \frac{\left(\prod_{i=1}^k z_i\right)^{t-1}}{\left(\sum_{i=1}^k z_i +1\right)^{t(k+1)+1}} \mathrm{d} z_1 \mathrm{d} z_2 \cdots \mathrm{d} z_k \end{align}

Note that for a single variable $z_1$, the integral $\int_0^1 \frac{z^{t-1}}{(1+z)^{2t+1}}\mathrm{d} z = 2F1 (t,1+2t,1+t,-1)/t$, that is, the Hypergeometic function.

Is this representation is helpful? Any ideas how to tackle this?

Any suggestions are much appreciated.

• Just to be sure, is your $\gamma(t,x)$ equal to $$\int_{0}^{x} u^{t-1}e^{-u}\,du$$ ? – Jack D'Aurizio May 25 '16 at 14:26
• Moreover, do you need a closed form or tight approximations / asymptotics for large values of $k$ are enough? I am asking this since the closed form is a hypergeometric $\phantom{}_2 F_1$ function even for $k=1$. – Jack D'Aurizio May 25 '16 at 14:34
• Thank you @JackD'Aurizio . Yes, the lower incomplete gamma is equal to $\int_0^x u^{t-1} e^{-u} \mathrm{d} u$ link. I am looking for a closed form (hypergeometric) series solution. Asymptotically, it can be shown that this is scales like $\log k + (t-1)\log \log k - log(t!)$, as $k \to \infty$. – sefi May 25 '16 at 15:07