Integration $\int_0^t x^{5n-1}\ e^{-x/a}\ \mathrm dx$ without Gamma function How to integrate this function without using the gamma function?
$$\int_0^t x^{5n-1}\ e^{-x/a}\ \mathrm dx$$
 A: Let $a=1/s$ and $5n-1=k$ so the integral looks more familiar:
$$
\int_0^t x^k e^{-sx}dx
$$
Now, I feel like I should integrate to infinity, so note that:
$$
\int_0^t x^k e^{-sx}dx = \int_0^\infty x^k e^{-sx}dx -\int_t^\infty x^k e^{-sx}dx
$$
The first integral is a well-known Laplace Transform
 case:
$$
\int_0^\infty x^k e^{-sx}dx = \frac{k!}{s^k}
$$
Now, I'd like to transform the second integral to start from $0$ as well, so I'll make the following change of variables:
$$
u = x-t \iff x = u+t
$$
Such that:
$$
\int_t^\infty x^k e^{-sx}dx = \int_0^\infty (u+t)^k e^{-su -st}du =e^{-st} \int_0^\infty (u+t)^k e^{-su}du
$$
Using binomial expansion:
$$
\int_t^\infty x^k e^{-sx}dx =e^{-st} \int_0^\infty (u+t)^k e^{-su}du = e^{-st}\int_0^\infty \sum _{j=0}^k u^{k-j}t^j \binom{j}{k} e^{-su}du 
$$
Thus:
$$
\int_t^\infty x^k e^{-sx}dx =  e^{-st}\sum _{j=0}^k \frac{(k-j)!}{s^{k-j}}t^j \binom{j}{k}
$$
Finally:
$$
\int_0^t x^k e^{-sx}dx = \frac{k!}{s^k} - e^{-st}\sum _{j=0}^k \frac{(k-j)!}{s^{k-j}}t^j \binom{j}{k}
$$
Now you need to replace back $a$ and $n$.
A: Assuming you mean $\int_0^x t^{2n-1}e^{-t/a}dt$, use integration by parts, 2n-1 times.  For the first step, let $u= t^{2n-1}$, $dv= e^{-t/a}dt$.  Then $du= (2n-1)t^{2n-2}dt$ and $v= -ae^{-t/a}$ so the integral becomes $\left[-at^{2n-1}e^{-t/a}\right]_0^x- a\int_0^x t^{2n-2}e^{-t/a}dt= -ax^{2n-1}e^{-x/a}- a\int_0^x t^{2n-2}e^{-t/a}dt$.  Repeat until the power of t in the integral is reduced to 0.
