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Does the following integral converge? $$\int_{-\infty}^{\infty} \frac{e^{-x}}{1+x^2}dx$$ If yes, then what can be the exact value? My intution is that it must converge but I am not able to prove how? So I tried to calculate because if we can get finite value then convergence is done. I made the substitution $ x=\tan{\theta}$ the integral becomes $$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} e^{-\tan\theta}d\theta.$$ But there also I'm not able to proceed further. Kindly help. Thank you!

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    $\begingroup$ Don't you think that there is a problem with large negative values of $x$ ? $\endgroup$ – Claude Leibovici Jan 15 '17 at 8:16
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Hint

Considering $$I=\int_{-\infty}^{\infty} \frac{e^{-x}}{1+x^2}dx=\int_{-\infty}^{0} \frac{e^{-x}}{1+x^2}dx+\int_{0}^{\infty} \frac{e^{-x}}{1+x^2}dx$$ For the first integral, change variable $x=-y$ which leads to $$I=\int_{0}^{\infty} \frac{e^{y}}{1+y^2}dy+\int_{0}^{\infty} \frac{e^{-x}}{1+x^2}dx$$ The second integral does not make any problem since $$\int_{0}^{\infty} \frac{e^{-x}}{1+x^2}dx<\int_{0}^{\infty} \frac{dx}{1+x^2}=\frac \pi 2$$ For the first integral, I am sure that you remember that $e^y$ varies much faster than any power of $y$.

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  • $\begingroup$ How to estimate first integral then ? it must diverge $\endgroup$ – Subhash Chand Bhoria Jan 15 '17 at 12:20
  • $\begingroup$ @SubhashChandBhoria; I am afraid that this is the case ! $\endgroup$ – Claude Leibovici Jan 15 '17 at 18:05

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