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Let us say we need to perform the classic integral $$ I=\int_{-\infty}^{+\infty}dz \,\frac{e^{itz}}{z^2+1}~, $$ where $t>0$.

What is normally done is the following. We consider the integral $$ K=\oint dz \,\frac{e^{itz}}{z^2+1} $$ with the contour closed on the positive imaginary part of the complex plane. In the area enclosed by the contour, we have the single pole $z_{pole}=+i$ and the rest of the function $f(z)=e^{itz}/(z+i)$ is holomorphic. By using the residue theorem, we can say $K=2\pi if(+i)=e^{-t}/\pi$. The integration over the arc goes to zero since the function $f(z)$ goes to zero faster than $O(1/z)$ for any $z$ in the contour. So, eventually we have $I=K=e^{-t}/\pi$. Fine.

What if I say: we consider the integral $$ T=\oint dz \,\frac{e^{itz}}{z^2+1} e^{-Im[z]} $$ with the same contour. Again, in the area enclosed by the contour, we have the single pole $z_{pole}=+i$ and the rest of the function $g(z)=e^{itz}e^{-Im[z]}/(z+i)$ is holomorphic. By using the residue theorem, I should be able to state that $T=2\pi ig(+i)=e^{-t-1}/\pi$. The integration over the arc goes again to zero since $Im[z]>0$ in the arc. Moreover, $Im[z]=0$ on the real axis, so we should have $I=T=e^{-t-1}/\pi$, which is different from before.

What did I do wrong?

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    $\begingroup$ why do you say that $g(z)=e^{itz}e^{-Im(z)}/(z+i)$ is holomorphic? $\endgroup$ – wisefool Apr 11 '15 at 21:24
  • $\begingroup$ @wisefool Isn't $g(z)$ analytic in the upper semicircle? It does not have any singularity and it looks to me differentiable in any point. Why shouldn't it be? $\endgroup$ – Wizzerad Apr 11 '15 at 21:59
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The function $\Im[z]$ is not analytic, since it does not satisfy Cauchy–Riemann equations. Therefore $e^{-\Im[z]}$ is also not analytic. The Cauchy formula cannot be applied and the second method is thus wrong. This solves the discrepancy. Finally, I thank @wisefool for his comment.

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  • $\begingroup$ Yes, that is what i was trying to point out in my comment! Sorry for my elapsed time of response, but this allowed you to figure it out by yourself! $\endgroup$ – wisefool Apr 13 '15 at 9:14

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