I wonder if there is any "exact" upper bound and lower bound for the following integrals:
$$ \int_0^1 \frac{e^{-\frac{x^2}{16(t-s)}}}{(t-s)^{m}}(1-s)^ads $$ Here $x\geq 0$, $t>1$, $m=\frac{1}{2}, 1, \frac{3}{2}$ and $0<a<\frac{1}{2}$.
By an "exact" bound, say $B(x)$, for a function $f(x)$ at $x=a$, means that the limit of the ratio $\frac{B(x)}{f(x)}$ as $x\rightarrow a$ is $1$.
Following attempts were made: First note that if $t=1$, then the integral can easily be reduced to a monomial multiple of an incomplete gamma function then some exact bounds could be made using the result from the article Exact lower and upper bounds on the incomplete gamma function" by Iosif Pinelis.
My main concern thus is when $t>1$. Consider the case $m=\frac{3}{2}$.
First for an upper bound, by some elementary inequalities involving the integrands, I was successful obtaining the following bound:
$$ \int_0^1 \frac{e^{-\frac{x^2}{16(t-s)}}}{(t-s)^{\frac{3}{2}}}(1-s)^ads\leq \min\left(e^{-\frac{x^{2}}{16t}}\int_{0}^{1}\frac{\left(1-s\right)^{a}}{\left(t-s\right)^{\frac{3}{2}}}ds,\int_0^1 \frac{e^{-\frac{x^2}{16(s+t-1)}}}{(s+t-1)^{\frac{3}{2}}}ds\right) $$
It is speculated that the first bound is exact at $t=\infty$, but not at $t=1$ as the ratio blows up to $\infty$. And the second bound is not exact both at $t=1, \infty$ but the ratios are finite. Also, from the bounds point of view, the second bound can be expressed as a difference of monomial multiples of some incomplete gamma functions, but the integral from the first bound require its own bound (which really corresponds to the case $x=0$ from our original integral).
To obtain an exact upper bound, I could take the minimum of the first bound and some scalar multiple of the second bound to ensure the "exactness" both at $t=1$ and $t=\infty$, but there is still lots of gaps to fill up to obtain a legit bound, including the problem of finding any possible "lower" bound.
I would like appreciate any comment for an improvement on this direction, or any other approach.