Dirichlet Integral/Proof 3

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Theorem

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


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:

\(\displaystyle \lim_{z \mathop \to 0} \frac {e^{i z} - 1} z\) \(=\) \(\displaystyle \intlimits {\frac {\map \d {e^{i z} } } {\d z} } {z = 0} {}\) Definition of Derivative of Complex Function
\(\displaystyle \) \(=\) \(\displaystyle i e^0\) Derivative of Exponential of a x
\(\displaystyle \) \(=\) \(\displaystyle i\)

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:

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

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:

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

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:

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

$\blacksquare$