Integral from $0$ to $t$ $\int \frac{e^x}{(t-x)^{9/10}} dx$

Question

What is the integral of $$\int_{0}^{t} \frac{e^x}{(t-x)^{\frac{9}{10}}}\; dx?$$

Let $u = t-x \Rightarrow dx = -du$

$$e^t \int_{0}^{t} \frac{e^{-u}}{u^{\frac{9}{10}}}\; du$$

How to proceed?

Answer 1

From what you did we have that

$$I = e^t \int_{0}^{t} \frac{e^{-u}}{u^{\frac{9}{10}}}\mathrm du = e^t\int_0^tu^{-9/10}e^{-u}\mathrm du = e^t\int_0^tu^{1/10-1}e^{-u}\mathrm du$$

If we use the definition of a Lower Incomplete Gamma Function

$$\gamma(s,t):= \int_0^tu^{s - 1}e^{-u}\mathrm du$$

So you have that your integral $I$ is

$$I = e^t\gamma(1/10,t)$$

We can use the relation

$$\gamma(s,x) = \Gamma(s) - \Gamma(s,x)$$

Then we will have that

$$I = I(t) = e^{t}\left(\Gamma(1/10)-\Gamma(1/10,t)\right)$$

where we use the definition

$$\Gamma(s,x) := \int_x^{+\infty}u^{s-1}e^{-u}\mathrm du$$

$$\Gamma(1/10) = 9.513507698668731836\dots $$

There is another approach. We say as before that $I(t) = e^t\gamma(1/10,t)$ and we can use that, for $a,x \in \mathbb{R}$ and $a>0,x>0$ we have

$$\gamma(a,x) = x^a\sum_0^{+\infty}\frac{(-1)^nx^n}{n!(a+n)}$$

Which gives you a series representation for your function of $t$. The proof that this series can be used is in here, what the link says is that this series can be created by termwise integration of

$$t^{a-1}e^{-t} = \sum_{0}^{+\infty}\frac{(-1)^n}{n!}t^{a+n-1}$$

And termwise integration is justified by uniform convergence of the power series for $e^{-t}$ on bounded intervals.