Dirichlet Integral

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Theorem

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


Proof 1

By Fubini's Theorem:

$\displaystyle \int_0^\infty \left({\int_0^\infty e^{- x y} \sin x \rd y}\right) \rd x = \int_0^\infty \left({\int_0^\infty e^{- x y} \sin x \rd x}\right) \rd y$


Then:

\(\displaystyle \int_0^\infty e^{- x y} \sin x \rd y\) \(=\) \(\displaystyle \left \lvert{- e^{- x y} \frac {\sin x} x}\right \rvert_0^\infty\) $\quad$ Primitive of $e^{a x}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac {\sin x} x\) $\quad$ $\quad$


and:

\(\displaystyle \int_0^\infty e^{- x y} \sin x \rd x\) \(=\) \(\displaystyle \left \lvert {\frac {-e^{- x y} \left({y \sin x + \cos x}\right)} {y^2 + 1} }\right \rvert_0^\infty\) $\quad$ Primitive of $e^{a x} \sin b x$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {y^2 + 1}\) $\quad$ $\quad$


Hence:

\(\displaystyle \int_0^\infty \frac {\sin x} x \rd x\) \(=\) \(\displaystyle \int_0^\infty \frac 1 {y^2 + 1} \rd y\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left \lvert{\arctan y}\right \rvert_0^\infty\) $\quad$ Primitive of $\dfrac 1 {x^2 + a^2}$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \frac \pi 2\) $\quad$ as $\displaystyle \lim_{y \mathop \to \infty} \arctan y = \frac \pi 2$ $\quad$

$\blacksquare$


Proof 2

Define:

$\displaystyle I \left({\alpha}\right) = \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$


Using Definite Integral of Partial Derivative:

\(\displaystyle I'\left(\alpha\right)\) \(=\) \(\displaystyle \frac \partial {\partial \alpha} \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \int_0^\infty \frac \partial {\partial \alpha} \frac {e^{-\alpha x} \sin x} x \rd x\) $\quad$ Leibniz Integral Rule $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\int_0^\infty e^{-\alpha x} \sin x \rd x\) $\quad$ Derivative of Exponential Function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left.{-\frac {e^{-\alpha x} \left({-\alpha \sin x + \cos x}\right)} {\left({-\alpha}\right)^2 + 1} }\right\rvert_0^\infty\) $\quad$ Primitive of $e^{\alpha x} \sin b x$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle -\frac 1 {\alpha^2 + 1}\) $\quad$ $\quad$


Integrating with respect to $\alpha$:

\(\displaystyle I \left({\alpha}\right)\) \(=\) \(\displaystyle -\int \frac 1 {\alpha^2 + 1} \rd \alpha\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle C - \arctan \alpha\) $\quad$ Primitive of $\dfrac 1 {x^2 + a^2}$ $\quad$


To determine $C$, take $\alpha \to \infty$:

\(\displaystyle \lim_{\alpha \mathop \to \infty} \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x\) \(=\) \(\displaystyle \lim_{\alpha \mathop \to \infty} \left({C - \arctan \alpha}\right)\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int_0^\infty \frac {0 \times \sin x} x\) \(=\) \(\displaystyle C - \frac \pi 2\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle C\) \(=\) \(\displaystyle \frac \pi 2\) $\quad$ $\quad$


Hence:

$\displaystyle \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x = \frac \pi 2 - \arctan \alpha$


Setting $\alpha = 0$ yields:

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

$\blacksquare$


Proof 3

Let:

$\displaystyle f \left({z}\right) = \frac {e^{i z} - 1} z$

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

$\displaystyle \operatorname{Im} \left({f \left({z}\right)}\right) = \frac {\sin z} z$

So:

$\displaystyle \operatorname{Im} \int_0^\infty \frac {e^{i x} - 1} x \rd x = \int_0^\infty \frac {\sin x} x \rd x$

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

Let $\Gamma_R = C_R \cup \left[{-R, R}\right]$.

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 $\lim_{z \to 0} f \left({z}\right)$ to be finite:

\(\displaystyle \lim_{z \to 0} \frac {e^{i z} - 1} z\) \(=\) \(\displaystyle \frac{\rd\left({ e^{i z} }\right)} {\rd z} \Bigg\vert_{z = 0}\) $\quad$ Definition of Derivative of Complex Function $\quad$
\(\displaystyle \) \(=\) \(\displaystyle i e^0\) $\quad$ Derivative of Exponential of a x $\quad$
\(\displaystyle \) \(=\) \(\displaystyle i\) $\quad$ $\quad$

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:

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

Therefore:

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

Evaluating the integral on the left hand side:

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

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:

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

$\blacksquare$


Source of Name

This entry was named for Johann Peter Gustav Lejeune Dirichlet.