$$\text{Let}\,\, \space I(a) = \int_{0}^{\frac{\pi}{4}} {e^{x} \tan^{a} x} \space .\space \text{Find}
\space \lim_{a\to\infty}aI(a). $$
Well this is an integral that I'm curently dealing with and so far I've tried to solve it using different approaches. Still, I haven't solved it and what I find always is an undefined answer. Even when I used mathematica it returned an undefined answer. This solution in my opinion is most likely to be the one that works: $$\\$$ Our Integral is $$\int_{0}^{\frac{\pi}{4}}e^x\space \lim_{a\to\infty}[{a\tan^a{x}}]\space dx,$$
letting $$A=\lim_{a\to\infty}{a\tan^a{x}}$$
We know that $$x \in [0,\frac{\pi}{4}]\space \Rightarrow \space \tan{x} \in[0,1] \space \Rightarrow \space \lim_{a\to\infty}a\tan{^ax}=
\ \begin{cases}
0 & 0\leq x< \frac{\pi}{4} \\
1 & x=\frac{\pi}{4}\\
\end{cases}
\
$$
Which tells us that
$$ A =
\ \begin{cases}
0 \times \infty & 0\leq x< \frac{\pi}{4} \\
\infty & x=\frac{\pi}{4}\\
\end{cases}
$$
Which are some undefined integrands… . Mathematica says the integral is undefined but the book, Advanced Calculus Explored, says it has an answer.
I think this is the right solution and I just need to work on the limit a little bit more, but this is something that I'm stuck in. I appreciate any kind of hints or helps.
Best Answer
Our problem is a disguised form of the result
$$\tag 1 \lim_{a\to \infty} a\int_0^1 y^a f(y)\,dy=f(1),$$
which holds for any continuous $f$ on $[0,1].$ It has been proved on MSE many times. The proof is quite simple for $f$ continuously differentiable on $[0,1],$ using integration by parts.
To see why the above will help us in the question at hand, let $x=\arctan y.$ The original expression then turns into
$$a\int_0^1 y^a e^{\arctan y}\frac{1}{1+y^2}\,dy.$$
Letting $f(y) = e^{\arctan y}\frac{1}{1+y^2},$ we see that $(1)$ implies our limit is
$$f(1)= e^{\arctan 1}\frac{1}{1+1^2} = \frac{e^{\pi/4}}{2}.$$