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.
On
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 $.
(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 $\frac1a + \frac{a}{1-b}$.