Compute $\int_{0}^{\infty}e^{-tz}(z+d)^{n-1}dz$ as a function of $\Gamma(n)$

Question

Is it possible to compute this integral $$\int_{0}^{\infty}e^{-tz}(z+d)^{n-1}dz$$ as a function of complete gamma $\Gamma(n)$. If possible, I'm looking for a closed form solution.

Thanks!

Answer 1

No, it is not possible to compute the integral as a function of $\Gamma(n)$

The integral is an Incomplete Gamma function. Change the variable $Z=t(z+d)$ and the integral definition of the Incomplete Gamma appears.

$\int_{0}^{\infty}e^{-tz}(z+d)^{n-1}dz = e^{td}t^{-n}\int_{td}^{\infty}e^{-Z}Z^{n-1}dZ = e^{td}t^{-n}\Gamma(n,td)$

Answer 2

A related technique. Here is a closed form for $n\in \mathbb{N}$. Recalling the binomial theorem and making the change of variables $u=tz$ we have $$I= \int_{0}^{\infty}e^{-tz}(z+d)^{n-1}dz = \int_{0}^{\infty}e^{-u}(u/t+d)^{n-1}\frac{du}{t} $$

$$ = \sum_{k=0}^{n-1} {n-1\choose k} \frac{d^{n-1-k}}{t^k} \int_{0}^{\infty}u^{k}e^{-u}du $$

$$ \implies I= d^{n-1}\sum_{k=0}^{n-1} {n-1\choose k} \frac{1}{(td)^k}\Gamma(k+1) $$

Added: If $n$ is not a natural number then you can consider Newton's generalised binomial theorem as stated in the comment by Cameron Williams.