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