Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Prove that $\displaystyle \lim_{n \to \infty} \int_0^\infty \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx$ exists and determine its value.

share|cite|improve this question
up vote 18 down vote accepted

Hint: Make the substitution $x=y/n$ and apply the dominated convergence theorem to see that

$$ \lim_{n \to \infty} \int_0^{\infty} \frac{e^{-x} \cos x}{nx^2+1/n}\,dx = \int_{0}^{\infty} \frac{dy}{y^2+1} = \frac{\pi}{2}. $$

share|cite|improve this answer
+1. Much shorter than mine. – Amr Jan 5 '13 at 1:59
Check this. – Mhenni Benghorbal Jan 6 '13 at 1:18

Hint: Break this integral into two parts. $$ \int_0^\infty \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx=\int_0^1 \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx+\int_1^\infty \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx$$ For the first part we use integration by parts to show that: $$\int_0^1 \frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}dx=[tan^{-1}(nx)e^{-x}cos(x)]^{1}_{0}-\int_0^1 tan^{-1}(nx)(e^{-x}\cos{x})'dx...(1)$$ The limit of the term $[tan^{-1}(nx)e^{-x}cos(x)]^{1}_{0}$ can be found easily. The limit of the integral $\int_0^1 tan^{-1}(nx)(e^{-x}\cos{x})'dx$ can be found by noting that $$\forall x\in[0,1][tan^{-1}(nx)(e^{-x}\cos{x})'\leq\frac{\pi}{4}(e^{-x}\cos{x})']$$ Finally, use Lesbesuge's dominated convergence to find the limit of the last integral apperaring in (1)

The limit of the second part can be found by noting that $\forall n\in Z^+\forall x\geq 1[x^2+1\leq nx^2+\frac{1}{n}]$. Thus: $$\forall n\in Z^+\forall x\geq1[\frac{e^{-x}\cos{x}}{nx^2 + \frac{1}{n}}\leq\frac{e^{-x}\cos{x}}{x^2 + 1}]$$ Now you can use Lebesuge's dominated convergence theorem to evaluate the second integral

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.