Computing a tricky limit $\lim_{n\to\infty} \sqrt{n}\int_0^{\infty} \cos^{2n-1}(x) e^{- \pi x} \ dx$ I'm interested in some neat approaches for
$$\lim_{n\to\infty} \sqrt{n}\int_0^{\infty} \cos^{2n-1}(x) e^{- \pi x} \ dx$$
Since I suspect my approach is wrong, and I don't wanna influence you in any way, I'll add it
in a comment  after the correct approach is posted. I also plan to ofer 100 bounty for the nicer,   simpler  approach. 

Since 
$$\int_{0}^{+\infty}\cos(n x)\,e^{-\pi x}\,dx = \frac{\pi}{n^2+\pi^2}$$
it is sufficient to compute the Fourier cosine series of $\cos^{2n-1}x.$ We have:
$$\cos^{2n-1}x = \frac{2}{4^n}(e^{ix}+e^{-ix})^{2n-1} = \frac{1}{4^{n-1}}\sum_{k=0}^{n-1}\binom{2n-1}{k}\cos((2n-2k-1)x)$$
hence:
$$\int_{0}^{+\infty}\cos^{2n-1}(x)\,e^{-\pi x}=\frac{1}{4^{n-1}}\sum_{k=1}^{n}\binom{2n-1}{n-k}\frac{\pi}{(2k-1)^2+\pi^2}.$$
Since for any fixed $k\geq 1$ we have:
$$ \lim_{n\to +\infty}\frac{\sqrt{n}}{4^{n-1}}\binom{2n-1}{n-k}=\frac{2}{\sqrt{\pi}}$$
the value of the limit equals:
$$ L = 2\sqrt{\pi}\sum_{k=1}^{+\infty}\frac{1}{\pi^2+(2k-1)^2} = \frac{\sqrt{\pi}}{2}\tanh\frac{\pi^2}{2},$$
where the last identity follows from considering the logarithmic derivatives of the Weierstrass product for the $\cosh$ function.
Notice that this problem in disguise already appeared here.
