Integration help!

  • Thread starter GregoryGr
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  • #1
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Homework Statement


Calculate the integral:

gif.latex?\int_{-\infty}^{0}&space;\frac{e^x}{1+cos^{2}(2x)}dx.gif


Homework Equations



-

The Attempt at a Solution



I tried some trig identities, like t=tan(x/2). The cos^2 smells like an arctan derivative but I can't seem to think of anything that could work...
 

Answers and Replies

  • #2
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You know cos^2(2x) is non-negative so it means 1+ cos^2(2x) is always positive
Use the fact that 1+cos^2(2x) is bounded to evaluate this integral .
 
Last edited:
  • #3
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Maybe write cos in exponents?
 
  • #4
Ray Vickson
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Homework Statement


Calculate the integral:

gif.latex?\int_{-\infty}^{0}&space;\frac{e^x}{1+cos^{2}(2x)}dx.gif


Homework Equations



-

The Attempt at a Solution



I tried some trig identities, like t=tan(x/2). The cos^2 smells like an arctan derivative but I can't seem to think of anything that could work...

I am beginning to doubt there is a simple closed-form solution. However, one can reduce it to a finite integration that might be preferable to use if you want an accurate numerical value. Call the integral J, and note that we can re-write it as an integral over [0,∞):
[tex] J = \int_0^{\infty} f(x) \, dx, \:\: f(x) = \frac{e^{-x}}{1 + \cos^2(2x)} [/tex]
Since ##\cos^2(2x)## is periodic with period ##\pi/2## we have
[tex] f\left( n \frac{\pi}{2} + t \right) = \alpha^n f(t), \; \alpha = e^{-\pi/2} [/tex]
so
[tex] J = \sum_{n=0}^{\infty} \alpha^n J_0 = \frac{J_0}{1-\alpha}, \text{ where }
J_0 = \int_0^{\pi/2} f(x) \, dx. [/tex]
For numerical work it might be better to work with J_0 instead of the original J.
 

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