Dirichlet Integral/Proof 3

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Theorem

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


Proof

Let:

$\map f x = \begin {cases} \dfrac {e^{i x} - 1} x & x \ne 0 \\ i & x = 0 \end {cases}$

We have, by Euler's Formula, for $x \in \R$:

$\map \Im {\map f x} = \begin {cases} \dfrac {\sin x} x & x \ne 0 \\ 1 & x = 0 \end {cases}$

So:

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

Let $C_R$ be the arc of the circle of radius $R$ centred at the origin connecting $R$ and $-R$ anticlockwise.

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

Then, by Contour Integral of Concatenation of Contours:

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

From Linear Combination of Contour Integrals, we write:

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

Note that $f$ is holomorphic inside our contour.

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

$\ds \oint_{\Gamma_R} \frac {e^{i x} - 1} x \rd x = 0$

We also have:

\(\ds \size {\int_{C_R} \frac { e^{i x} } x \rd x}\) \(\le\) \(\ds \pi \max_{0 \mathop \le \theta \mathop \le \pi} \size {\frac 1 {R e^{i \theta} } }\) Jordan's Lemma
\(\ds \) \(=\) \(\ds \frac \pi R\)
\(\ds \) \(\to\) \(\ds 0\) as $R \to \infty$

Therefore:

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

Evaluating the integral on the left hand side:

\(\ds \int_{C_R} \frac {\rd x} x\) \(=\) \(\ds \int_0^\pi \frac {i R e^{i \theta} } {R e^{i \theta} } \rd \theta\) Definition of Complex Contour Integral
\(\ds \) \(=\) \(\ds i \int_0^\pi \rd \theta\)
\(\ds \) \(=\) \(\ds \pi i\)

So:

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

Taking the imaginary part:

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

From Definite Integral of Even Function:

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

Hence:

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

$\blacksquare$