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

• MHB
• lfdahl
In summary, the limit of the integral challenge of (e^(-x)cosx)/(1/n+nx^2) is 0 and can be solved using the substitution method, integration by parts, partial fraction decomposition, or trigonometric identities. The value of n affects the limit, as it approaches infinity the limit approaches 0. This limit is significant in determining the convergence of the integral, with a limit of 0 indicating convergence and any other value indicating divergence.
lfdahl
Gold Member
MHB
Find

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

Nice one.
My attempt:
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}

Klaas van Aarsen said:
Nice one.
My attempt:
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:
$I = \lim_{n \to \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{\frac{1}{n}+nx^2}dx = \lim_{n \to \infty}\int_{0}^{\infty}\frac{e^{-x}\cos x}{1+n^2x^2}ndx = \lim_{n \to \infty}\int_{0}^{\infty}\frac{e^{-u/n}\cos (u/n)}{1+u^2}du$

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

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

## 1. What is the limit of the given integral challenge?

The limit of the integral challenge is equal to 0.

## 2. How do you solve for the limit of the given integral challenge?

To solve for the limit, we can use the properties of limits and the fundamental theorem of calculus. We can also use L'Hopital's rule to simplify the expression and then evaluate the limit.

## 3. Why is the limit of the given integral challenge equal to 0?

The limit is equal to 0 because as x approaches infinity, the numerator (e^(-x)cosx) approaches 0 while the denominator (1/n + nx^2) approaches infinity. This results in the overall expression approaching 0.

## 4. Can the limit of the given integral challenge be evaluated using substitution?

No, substitution cannot be used to evaluate the limit in this case. The expression contains both a variable in the numerator and the denominator, and substitution is only applicable when the variable is in the numerator.

## 5. How does the value of n affect the limit of the given integral challenge?

The value of n affects the limit by changing the rate at which the denominator approaches infinity. As n increases, the denominator increases at a faster rate, resulting in the limit approaching 0 faster.

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