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Show that $\displaystyle\int_0^1e^{t-1}t^x(1+t^2)^{-x}\mathrm dt\sim\left(\frac{\pi}{2x}\right)^{1/2}2^{-x}$ to leading order as $x\to\infty$.

I'm working old qualifying exam problems, and I can't seem to figure out how to manipulate this one into a useable form. I'm assuming there is some substitution after getting this into the form

$\displaystyle\int_0^1e^{t-1+x\ln(t)-x\ln(1+t^2)}\mathrm dt$,

which could make the rest straightforward.

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    Given the title, I suppose you know this is simple to solve through Laplace's method: https://en.wikipedia.org/wiki/Laplace's_method – 2017-01-18
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    Note that $f(t)=\frac{t}{1+t^2}$ goes monotonically from $0$ to $1/2$ as $t$ goes from $0$ to $1$. You then have $f(t)^x g(t)$ where $g$ ranges between two fixed positive numbers. Rewrite $f(t)^x$ as $e^{x \ln(f(t))}$. Where is $\ln(f(t))$ largest on $[0,1]$? Your Laplace method will see only that point. The only thing that is not literally handled by the Wikipedia article is that that point is an endpoint. – 2017-01-18

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