Dirichlet Integral/Proof 3

Proof
Let:


 * $\map f z = \dfrac {e^{i z} - 1} z$

We have, by Euler's Formula, on the real line:


 * $\map \Im {\map f z} = \dfrac {\sin z} z$

So:


 * $\displaystyle \map \Im {\int_0^\infty \dfrac {e^{i x} - 1} x \rd x} = \int_0^\infty \dfrac {\sin x} x \rd x$

Let $C_R$ be the semicircular contour of radius $R$ situated on the upper half plane, centred at the origin, traversed anti-clockwise.

Let $\Gamma_R = C_R \cup \closedint {-R} R$.

Then, by Contour Integral of Concatenation of Contours:


 * $\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = \int_{C_R} \frac {e^{i z} - 1} z \rd z + \int_{-R}^R \frac {e^{i x} - 1} x \rd x$

From Linear Combination of Contour Integrals, we write:


 * $\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = \int_{C_R} \frac {e^{i z} } z \rd z - \int_{C_R} \frac {\rd z} z + \int_{-R}^R \frac {e^{i x} - 1} x \rd x$

The only singularity of $f$ is at $z = 0$. However, we can show $\lim_{z \mathop \to 0} \map f z$ to be finite:

So, $z = 0$ is a removable singularity of $f$.

Therefore, $f$ is holomorphic inside our contour.

It then follows from the Cauchy-Goursat Theorem, that:


 * $\displaystyle \oint_{\Gamma_R} \frac {e^{i z} - 1} z \rd z = 0$

We also have:

Therefore:


 * $\displaystyle \lim_{R \mathop \to \infty} \int_{C_R} \frac {\rd z} z = \lim_{R \mathop \to \infty} \int_{-R}^R \frac {e^{i x} - 1} x \rd x = \int_{-\infty}^\infty \frac {e^{i x} - 1} x \rd x$

Evaluating the integral on the left hand side:

So:


 * $\displaystyle \int_{-\infty}^\infty \frac {e^{i x} - 1} x \rd x = \pi i$

Taking the imaginary part:


 * $\displaystyle \int_{-\infty}^\infty \frac {\sin x} x \rd x = \pi$

From Definite Integral of Even Function:


 * $\displaystyle \int_{-\infty}^\infty \frac {\sin x} x \rd x = 2 \int_0^\infty \frac {\sin x} x \rd x$

Hence:


 * $\displaystyle \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$