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I have to show that $$\int_0^{\infty} \frac{\cos(x)-e^{-x}}{x} dx = 0 $$ using a quarter circle in the upper positive plane and the function $f(z) = {e^{iz}}/{z}$.

I think I have a solution where I take a quarter-annulus of big radius $R$ and little radius $r$ on the quarter circle, and show they both tend to $0$ as $R \rightarrow \infty $ and $r \rightarrow 0$. However, I'm not sure how to manipulate the function $f$ to add in the $-e^{-x}$ term. I also know that $f(z) = 0$ by Cauchy's theorem.

Thus far I think I have a solution with the function $ f(z) = \frac{e^{iz}-e^{-x}}{z} $ but I am not sure this is the correct way of tackling the problem.

Connor Harris
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walle_whale
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  • Perhaps you can use that $\cos(x)=\frac{e^{ix}+e^{-ix}}{2}$ – Rhys Hughes May 14 '18 at 14:46
  • In the first quadrant, your integrand gets arbitrarily large in absolute value as $|z| \to \infty$ (a consequence of the $e^{-iz}$ term in the complex definition of $\cos$). I'm not sure that approach will be easy. – Connor Harris May 14 '18 at 15:26
  • Though I have a sneaking suspicion that an integral over a semi-circular contour in the upper half-plane may allow some of the divergent integrals to cancel each other. – Connor Harris May 14 '18 at 15:29
  • See here https://math.stackexchange.com/questions/217263/int-0-infty-frac-cos-x-e-xx-mathrm-dx-evaluate-integral?noredirect=1&lq=1 – cgiovanardi May 14 '18 at 16:40
  • Or use the Cantarini-Frullani's theorem: https://math.stackexchange.com/questions/1807193/evaluating-the-integral-of-frac-cosx-e-xx-using-contour-integratio – cgiovanardi May 14 '18 at 16:48

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