Bound for $\int_{0}^{t}e^{-xa^2}a(t-x)^{-b}\,dx$ I'm trying to estimate the following integral: $$\int_{0}^{t}e^{-xa^2}a(t-x)^{-b}\,dx$$ where constants $t,a>0$ and $0<b<1/2$. I want to get a bound for this integral. The preferred bound is like:$f(t)/a$, with $f(t)\rightarrow C$ as $t\rightarrow \infty$, where C is a constant. For example, $f(t)=1-e^{-ta^2}$ is good enough.
I tried integration by parts but failed.
 A: (Edit: This doesn't solve the problem!)
This is a crude approximation, but it gives what you ask for. 
We have $(t-x)^{-b} \ge 1$ for $x\ge t-1$ and $(t-x)^{-b}<1$ for $x < t-1$. Therefore:
$$ 
\begin{split}
\int_0^t e^{-xa^2}a(t-x)^{-b}\, dx
&\le \int_0^{t-1} ae^{-xa^2} \, dx + \int_{t-1}^{t}a(t-x)^{-b} \, dx \\
&= \frac{1}{a}\left(1-e^{(1-t)a^2}\right) + \frac{a}{1-b} 
\end{split} 
$$
which has limit $\frac1a + \frac{a}{1-b}$. 
A: I get
$\begin{array}\\
\int_{0}^{t}e^{-xa^2}a(t-x)^{-b}\,dx
&=a^{2b-1}e^{-ta^2}\int_{0}^{ta^2}e^{y}y^{-b}dy\\
&=a^{2b-1}e^{-ta^2}\gamma(-b+1, ta^2)\\
&<a^{2b-1}e^{-ta^2}\Gamma(-b+1)\\
&\le a^{2b-1}e^{-ta^2}\Gamma(1/2)\\
&\le \sqrt{\pi}a^{2b-1}e^{-ta^2}\\
\end{array}
$
where
$\gamma(,)$ is the incomplete gamma function.
Here's my steps.
$\begin{array}\\
I(a, b, t)
&=\int_{0}^{t}e^{-xa^2}a(t-x)^{-b}dx\\
&=a\int_{0}^{t}e^{-(t-x)a^2}x^{-b}dx\\
&=a\int_{0}^{t}e^{-ta^2}e^{xa^2}x^{-b}dx\\
&=ae^{-ta^2}\int_{0}^{t}e^{xa^2}x^{-b}dx\\
&=ae^{-ta^2}\int_{0}^{ta^2}e^{y}(y/a^2)^{-b}dy/a^2
\quad y=xa^2, x=y/x^2, dx = dy/a^2\\
&=ae^{-ta^2}a^{2b-2}\int_{0}^{ta^2}e^{y}y^{-b}dy\\
&=a^{2b-1}e^{-ta^2}\int_{0}^{ta^2}e^{y}y^{-b}dy\\
&=a^{2b-1}e^{-ta^2}\gamma(-b+1, ta^2)
\qquad\text{incomplete gamma function}\\
&<a^{2b-1}e^{-ta^2}\int_{0}^{\infty}e^{y}y^{-b}dy\\
&=a^{2b-1}e^{-ta^2}\Gamma(-b+1)\\
\end{array}
$
$0 < b < \frac12
\implies
\frac12 < -b+1 < 1
$.
Therefore
$\Gamma(-b+1)
\le \Gamma(1/2)
=\sqrt{\pi}
\approx 1.772
$.
