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I'd like some help solving the integral

$$ \int_{-\infty}^{\infty} e^{-2 \, \alpha \, |x|} \cdot \cos^2(x) \; \, dx $$

with $\alpha > 0$

I just assumed 'integration-by-parts' was the way to go, but the first part of the product alone ($e^{-2\alpha|x|}$) gets quite confusing. Is there any trick to making it stay manageable?

The absolute-value in the exponent is confusing me, too - do I have to differentiate between cases every time?

colorfly
  • 33

2 Answers2

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We have: $$ I = \int_{-\infty}^{+\infty}e^{-2\alpha|x|}\cos^2(x)\,dx = 2\int_{0}^{+\infty}e^{-2\alpha x}\cos^2 x\,dx =\int_{0}^{+\infty}e^{-2\alpha x}(1+\cos(2x))\,dx$$ hence: $$ I = \frac{1}{2\alpha}+\frac{\alpha}{2(\alpha^2+1)}$$ since: $$\int_{0}^{+\infty}e^{-\beta x}\,dx = \frac{1}{\beta},\qquad \int_{0}^{+\infty}e^{-\beta x}\cos x\,dx = \frac{\beta}{1+\beta^2}.$$

Jack D'Aurizio
  • 353,855
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Since it is an even function, you can only do it for x > 0. Cos^2 ( x ) = ( 1 - cos2x )/ 2, and exp( -2ax )* cos (2x) is the real part of exp( -2ax + 2ix ). Remain is easy.

Sky
  • 357