Saturday 12 April 2014

integration - What did I do wrong in calculating $int_{-infty}^{infty}frac{sin{x}}{x}mathrm{d}x$ via $oint_{gamma}frac{sin{z}}{z}mathrm{d}z$?

I tried to calculate the $\int_{-\infty}^{\infty}\frac{\sin{x}}{x}\mathrm{d}x$ by the complex integral $\oint_{\gamma}\frac{\sin{z}}{z}\mathrm{d}z$, where $\gamma$ is the half-circle with $y > 0$.



$$\begin{align*}

\int_{-\infty}^{\infty}\frac{\sin{x}}{x}\mathrm{d}x &= \Im\left \{\oint_{\gamma}\frac{\mathrm{e}^{iz}}{z}\mathrm{d}z \right \}=\Im\left \{ 2\pi i Res\left [ \frac{\mathrm{e}^{iz}}{z}, 0 \right ] \right \}=\Im\left \{ 2\pi i \lim_{z\to0} \left ( z \frac{\mathrm{e}^{iz}}{z} \right ) \right \}
\\ &= \Im\left \{ 2\pi i \lim_{z\to0} \mathrm{e}^{iz} \right \}=\Im\left \{ 2\pi i {e}^{i0} \right \}=\Im\left \{ 2\pi i \cdot 1 \right \}=2\pi
\end{align*}$$
Now, I know the integral $\int_{-\infty}^{\infty}\frac{\sin{x}}{x}\mathrm{d}x = \pi$ by many other theorems.
So, what did I do wrong?



If you know, can you please write the correct answer? Thanks!

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