MHB Limit of integral challenge of (e^(-x)cosx)/(1/n+nx^2)

[lfdahl]

Find

\[\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}dx.\]

 

[I like Serena]

Nice one.
My attempt:

Spoiler

Partial integration gives us:
\begin{aligned}\int\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}\,dx
&= \int e^{-x}\cos x\cdot \frac{n}{1+(nx)^2}\,dx \\
&= \int e^{-x}\cos x\,d\big(\arctan(nx)\big) \\
&= e^{-x}\cos x\arctan(nx) - \int\arctan(nx)\,d\big(e^{-x}\cos x\big)
\end{aligned}
Therefore:
\begin{aligned}\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}\,dx
&= \lim_{n\rightarrow \infty}\lim_{a\to 0^+,b\to\infty}\left[e^{-x}\cos x\arctan(nx)\Big|_a^b - \int_a^b\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{n\rightarrow \infty}\lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\lim_{n\rightarrow \infty}\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\frac\pi 2\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \frac\pi 2\,e^{-x}\cos x\right]_a^b \\
&=\frac \pi 2
\end{aligned}

 

[lfdahl]

Nice one.
My attempt:

Spoiler

Partial integration gives us:
\begin{aligned}\int\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}\,dx
&= \int e^{-x}\cos x\cdot \frac{n}{1+(nx)^2}\,dx \\
&= \int e^{-x}\cos x\,d\big(\arctan(nx)\big) \\
&= e^{-x}\cos x\arctan(nx) - \int\arctan(nx)\,d\big(e^{-x}\cos x\big)
\end{aligned}
Therefore:
\begin{aligned}\lim_{n\rightarrow \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}\,dx
&= \lim_{n\rightarrow \infty}\lim_{a\to 0^+,b\to\infty}\left[e^{-x}\cos x\arctan(nx)\Big|_a^b - \int_a^b\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{n\rightarrow \infty}\lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\lim_{n\rightarrow \infty}\arctan(nx)\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \int_a^b\frac\pi 2\,d\big(e^{-x}\cos x\big)\right] \\
&= \lim_{a\to 0^+,b\to\infty}\left[ - \frac\pi 2\,e^{-x}\cos x\right]_a^b \\
&=\frac \pi 2
\end{aligned}

Thankyou, Klaas van Aarsen, for your participation and a correct answer! (Yes)An alternative solution:

Spoiler

\[I = \lim_{n \to \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}dx = \lim_{n \to \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{1+n^2x^2}ndx = \lim_{n \to \infty}\int_{0}^{\infty}\frac{e^{-u/n}\cos (u/n)}{1+u^2}du\]

Since the functions $\frac{e^{-u/n}\cos (u/n)}{1+u^2}$ are boundedly convergent for all $u$ and $n$ between $0$ and infinity, we may use the Bounded Convergence Theorem to move the limit inside the integral giving

\[I =\int_{0}^{\infty}\frac{1}{1+u^2}du = \frac{\pi}{2}.\]