# Dirichlet Integral

## Theorem

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

## Proof 1

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

Then:

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

and:

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

Hence:

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

$\blacksquare$

## Proof 2

Define:

$\displaystyle \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$
 $\displaystyle \map {I'} \alpha$ $=$ $\displaystyle \frac \partial {\partial \alpha} \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$ $\displaystyle$ $=$ $\displaystyle \int_0^\infty \frac \partial {\partial \alpha} \frac {e^{-\alpha x} \sin x} x \rd x$ Leibniz's Integral Rule $\displaystyle$ $=$ $\displaystyle -\int_0^\infty e^{-\alpha x} \sin x \rd x$ Derivative of Exponential Function $\displaystyle$ $=$ $\displaystyle \intlimits {-\frac {e^{-\alpha x} \paren {-\alpha \sin x + \cos x} } {\paren {-\alpha}^2 + 1} } 0 \infty$ Primitive of $e^{\alpha x} \sin b x$ $\displaystyle$ $=$ $\displaystyle -\frac 1 {\alpha^2 + 1}$

Integrating with respect to $\alpha$:

 $\displaystyle \map I \alpha$ $=$ $\displaystyle -\int \frac 1 {\alpha^2 + 1} \rd \alpha$ $\displaystyle$ $=$ $\displaystyle C - \arctan \alpha$ Primitive of $\dfrac 1 {x^2 + a^2}$

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

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

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:

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

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

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

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:

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

Hence:

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

$\blacksquare$

## Proof 4

$\displaystyle \int_0^\infty {\dfrac {\map f x} x} = \int_0^{\to \infty} \map F u \rd u$

for a real function $f$ and its Laplace transform $\laptrans f = F$, provided they exist.

Let $\map f x := \sin x$.

Then from Laplace Transform of Sine:

$\laptrans {\map f x} = \dfrac 1 {s^2 + 1}$

Hence:

 $\displaystyle \int_0^\infty \frac {\sin x} x \rd x$ $=$ $\displaystyle \int_0^{\to \infty} \dfrac {\d u} {u^2 + 1}$ $\displaystyle$ $=$ $\displaystyle \bigintlimits {\arctan u} 0 \infty$ Primitive of $\dfrac 1 {x^2 + a^2}$ $\displaystyle$ $=$ $\displaystyle \dfrac \pi 2$

$\blacksquare$

## Source of Name

This entry was named for Johann Peter Gustav Lejeune Dirichlet.