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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.

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2 Answers 2

up vote 18 down vote accepted

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

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

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+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

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