Dirichlet Integral

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Theorem

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


Proof 1

By Fubini's Theorem:

$\ds \int_0^\infty \paren {\int_0^\infty e^{- x y} \sin x \rd y} \rd x = \int_0^\infty \paren {\int_0^\infty e^{- x y} \sin x \rd x} \rd y$


Then:

\(\ds \int_0^\infty e^{- x y} \sin x \rd y\) \(=\) \(\ds \intlimits {-e^{- x y} \frac {\sin x} x} 0 \infty\) Primitive of $e^{a x}$
\(\ds \) \(=\) \(\ds \frac {\sin x} x\)


and:

\(\ds \int_0^\infty e^{- x y} \sin x \rd x\) \(=\) \(\ds \intlimits {\frac {-e^{- x y} \paren {y \sin x + \cos x} } {y^2 + 1} } 0 \infty\) Primitive of $e^{a x} \sin b x$
\(\ds \) \(=\) \(\ds \frac 1 {y^2 + 1}\)


Hence:

\(\ds \int_0^\infty \frac {\sin x} x \rd x\) \(=\) \(\ds \int_0^\infty \frac 1 {y^2 + 1} \rd y\)
\(\ds \) \(=\) \(\ds \bigintlimits {\arctan y} 0 \infty\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac \pi 2\) as $\ds \lim_{y \mathop \to \infty} \arctan y = \frac \pi 2$

$\blacksquare$


Proof 2

$\ds \int_0^\infty \frac {\sin x} x \rd x$ is convergent as an improper integral.

Indeed, for all $n \in \N$:

\(\ds \int_0^{2\pi n}\frac {\sin x} {x} \rd x\) \(=\) \(\ds \sum_{k \mathop = 0}^{2 n \mathop - 1}\int_{\pi k}^{\pi \paren {k + 1} }\frac {\sin x} {x} \rd x\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{2 n \mathop - 1} {\paren {-1} }^k \int_0^\pi \frac {\sin x} {x + \pi k} \rd x\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^{2 n \mathop - 1} \frac { {\paren {-1} }^k} {\pi k} \int_0^\pi \frac {\sin x} {1 + \frac x {\pi k} } \rd x\)

But:

\(\ds \size {\int_0^\pi \frac {\sin x} {1 + \frac x {\pi k} } \rd x - 2}\) \(\le\) \(\ds \int_0^\pi \sin x \size {\frac 1 {1 + \frac x {\pi k} } - 1} \rd x\)
\(\ds \) \(=\) \(\ds \frac 1 {k \pi} \int_0^\pi x \sin x \rd x\)

so that:

$\ds \int_0^\pi \frac {\sin x} {1 + \frac x {k \pi} } \rd x \to_{k \mathop \to \infty} 2$

Hence:

$\ds \int_0^{2\pi n }\frac {\sin x} x \rd x = \sum_{k \mathop = 0}^{n \mathop -1} \frac 1 {2 \pi k} \int_0^\pi \frac {\sin x} {1 + \frac x {2 \pi k} } \rd x - \frac 1 {\pi \paren {2 k + 1} } \int_0^\pi \frac {\sin x} {1 + \frac x {\pi \paren {2 k + 1} } } \rd x$

can be expressed as a series whose general term is equivalent to:

$\dfrac 2 \pi \times \dfrac 1 {2 k \paren {2 k + 1} }$

which is the term of an absolutely convergent series.


By Modulus of Sine of x Less Than or Equal To Absolute Value of x:

$\ds \size {\frac {e^{-\alpha x} \sin x} x} \le e^{-\alpha x}$

From Laplace Transform of Real Power:

$\ds \int_0^\infty e^{-\alpha x} \rd x = \frac 1 \alpha$

Hence by Comparison Test for Improper Integral:

$\ds \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$

converges whenever $\alpha > 0$.

So, we can define a real function $I : \openint 0 \infty \to \R$ by:

$\ds \map I \alpha = \int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x$

for each $\alpha \in \openint 0 \infty$.


Using Improper Integral of Partial Derivative on segments included in $\openint 0 \infty$:

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

Therefore, by Derivative of Arctangent Function

$\map I \alpha = -\arctan \alpha + K$

for some $K \in \R$.

We also have:

\(\ds \size {\map I \alpha}\) \(=\) \(\ds \size {\int_0^\infty \frac {e^{-\alpha x} \sin x} x \rd x}\)
\(\ds \) \(\le\) \(\ds \int_0^\infty \size {\frac {e^{-\alpha x} \sin x} x} \rd x\) Triangle Inequality for Definite Integrals
\(\ds \) \(\le\) \(\ds \frac 1 \alpha\)

so:

$\ds \lim_{\alpha \mathop \to \infty} \size {\map I \alpha} = 0$

That is:

$\ds \lim_{\alpha \mathop \to \infty} \map I \alpha = 0$

Therefore:

$\map I \alpha = \dfrac \pi 2 - \arctan \alpha$

since $\ds \arctan \alpha \to_{\alpha \mathop \to \infty} \frac \pi 2$.

Note that we have:

$\ds \map I \alpha \to_{\alpha \mathop \to 0} \frac \pi 2$

We now need to show that:

$\ds \map I \alpha \to_{\alpha \mathop \to 0} \int_0^\infty \frac {\sin x} x \rd x$

Observe for this purpose that:

\(\ds \map I \alpha\) \(=\) \(\ds \int_0^\infty \frac {\sin 2 x} x e^{-2 \alpha x} \rd x\) change of variable $x \mapsto 2 x$
\(\ds \) \(=\) \(\ds 2 \int_0^\infty \frac {\sin x} x e^{-2 \alpha x} \cos x \rd x\)
\(\ds \) \(=\) \(\ds 2 \intlimits {\frac {\sin^2 x} x e^{-2 \alpha x} } 0 \infty - 2 \int_0^\infty \sin x e^{-2 \alpha x} \paren {-\alpha \frac {\sin x} x + \frac {\cos x} x - \frac {\sin x} {x^2} } \rd x\) integration by Parts for improper integral
\(\ds \) \(=\) \(\ds 2 \alpha \int_0^\infty \frac {\sin^2 x} x e^{-2 \alpha x} \rd x + 2\int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2 \alpha x} \rd x - \map I \alpha\) by Continuous Real Function/Examples, $\dfrac {\sin x} x \to 1$ in $0$

where all the improper integrals appearing here are convergent by Comparison Test for Improper Integral, as used above for defining $\map I \alpha$.

Therefore:

$\ds \map I \alpha = \alpha \int_0^\infty \frac {\sin^2 x} x e^{-2 \alpha x} \rd x + \int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2 \alpha x} \rd x$


We also have:

\(\ds \int_0^\infty \frac {\sin x} x \rd x\) \(=\) \(\ds 2 \int_0^\infty \frac {\sin x} x \cos x \rd x\) by change of variable $x \mapsto 2 x$
\(\ds \) \(=\) \(\ds 2 \intlimits {\frac {\sin^2 x} x} 0 \infty - 2\int_0^\alpha \frac {\sin x} x \cos x \rd x + 2\int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x\)
\(\ds \) \(=\) \(\ds -\int_0^\infty \frac {\sin x} x \rd x + 2 \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x\)

where the improper integrals on the right hand side are convergent because the first one identifies with $\ds \int_0^\infty \frac {\sin x} x \rd x$ and the second one because $\dfrac {\sin^2 x} {x^2}$ is integrable on $\openint 0 \infty$, since it has a finite limit at $0$ and is smaller than $\frac 1 {x^2}$ at $\infty$.

Hence:

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

where the second integral is absolutely convergent.

Moreover:

\(\ds \alpha \int_0^\infty \frac {\sin^2 x} x e^{-2 \alpha x} \rd x\) \(=\) \(\ds \alpha \int_0^\infty \frac {\sin^2 \frac x \alpha} x e^{-2 x} \rd x\)
\(\ds \) \(\le\) \(\ds \alpha \paren {\int_0^\alpha \frac {\frac {x^2} {\alpha^2} } x \rd x + \int_\alpha^1 \frac 1 x \rd x + \int_1^\infty e^{-2 x} \rd x}\)
\(\ds \) \(=\) \(\ds \alpha \paren {\frac 1 2 - \ln \alpha + \frac 1 {2 e^2} }\)
\(\ds \) \(\to_{\alpha \mathop \to 0}\) \(\ds 0\)

whenever $\alpha \le 1$.

Also:

$\ds \int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2\alpha x} \rd x \to_{\alpha\to 0} \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x$

This is because, for any positive $R$ and $\alpha$:

\(\ds \int_0^\infty {\paren {\frac {\sin x} x} }^2 \paren {1 - e^{-2 \alpha x} }\rd x\) \(=\) \(\ds \int_0^{\frac 1 {\sqrt \alpha} } {\paren {\frac {\sin x} x} }^2 \paren {1 - e^{-2 \alpha x} }\rd x + \int_{\frac 1 {\sqrt \alpha} }^\infty {\paren {\frac {\sin x} x} }^2 \paren {1 - e^{-2 \alpha x} }\rd x\)
\(\ds \) \(\le\) \(\ds \paren {1 - e^{-2 \sqrt \alpha} } \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x + \int_{\frac 1 {\sqrt \alpha} }^\infty {\paren {\frac {\sin x} x} }^2 \rd x\)
\(\ds \) \(\to_{\alpha \mathop \to 0}\) \(\ds 0\)

because $\dfrac {\sin^2 x} {x^2}$ is integrable on $\openint 0 \infty$.


Finally, we have:

\(\ds \map I \alpha\) \(=\) \(\ds \alpha \int_0^\infty \frac {\sin^2 x} x e^{-2 \alpha x} \rd x + \int_0^\infty {\paren {\frac {\sin x} x} }^2 e^{-2 \alpha x} \rd x\)
\(\ds \) \(\to_{\alpha \mathop \to 0}\) \(\ds \int_0^\infty {\paren {\frac {\sin x} x} }^2 \rd x\)
\(\ds \) \(=\) \(\ds \int_0^\infty \frac {\sin x} x \rd x\)

as well as:

\(\ds \map I \alpha\) \(=\) \(\ds \frac \pi 2 - \arctan \alpha\)
\(\ds \) \(\to_{\alpha \mathop \to 0}\) \(\ds \frac \pi 2\)

So that, by uniqueness of limits:

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

$\blacksquare$


Proof 3

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:

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

$\blacksquare$


Proof 4

From Integral to Infinity of Function over Argument:

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

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:

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

$\blacksquare$


Proof 5

Let $M \in \R_{>0}$.

Define a real function $I_M : \R \to \R$ by:

$\ds \map {I_M} \alpha := \int_0^M \dfrac {\sin x} x e^{-\alpha x} \rd x$

Then, for $\alpha > 0$:

\(\ds \size {\map {I_M} \alpha}\) \(\le\) \(\ds \int_0^M \size {\dfrac {\sin x} x e^{-\alpha x} } \rd x\) Absolute Value of Definite Integral
\(\ds \) \(\le\) \(\ds \int_0^M e^{-\alpha x} \rd x\) Sine Inequality $\size {\sin x} \le \size x$
\(\ds \) \(=\) \(\ds \intlimits {\dfrac {e^{-\alpha x} } {-\alpha} } 0 M\) Primitive of $e^{ax}$
\(\ds \) \(=\) \(\ds \dfrac 1 \alpha - \dfrac {e^{-\alpha M} } \alpha\)
\(\text {(1)}: \quad\) \(\ds \) \(\le\) \(\ds \dfrac 1 \alpha\)


On the other hand:

\(\ds \map {I'_M} \alpha\) \(=\) \(\ds \int_0^M \dfrac \partial {\partial \alpha} \paren {\dfrac {\sin x} x e^{-\alpha x} } \rd x\) Definite Integral of Partial Derivative
\(\ds \) \(=\) \(\ds \int_0^M - \sin x e^{-\alpha x} \rd x\) Primitive of $e^{ax}$
\(\ds \) \(=\) \(\ds \intlimits {-\frac {e^{-\alpha x} \paren {-\alpha \sin x + \cos x} } {\paren {-\alpha}^2 + 1} } 0 M\) Primitive of $e^{\alpha x} \sin b x$
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {-1} {\alpha^2 + 1} + \cos M \dfrac {e^{-\alpha M} }{\alpha^2 + 1} + \sin M \dfrac {\alpha e^{-\alpha M} }{\alpha^2 + 1}\)

Thus:

\(\ds \map {I_M} A - \map {I_M} 0\) \(=\) \(\ds \int_0^A \map {I'_M} \alpha \rd \alpha\) Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds - \int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} + \cos M \int_0^A \dfrac {e^{-\alpha M} }{\alpha^2 + 1} \rd \alpha + \sin M \int_0^A \dfrac {\alpha e^{-\alpha M} }{\alpha^2 + 1} \rd \alpha\) by $\paren 2$ and Linear Combination of Integrals

Thus:

\(\ds \size {\map {I_M} A - \map {I_M} 0 + \int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} }\) \(=\) \(\ds \size {\cos M \int_0^A \dfrac {e^{-\alpha M} }{\alpha^2 + 1} \rd \alpha + \sin M \int_0^A \dfrac {\alpha e^{-\alpha M} }{\alpha^2 + 1} \rd \alpha}\)
\(\ds \) \(\le\) \(\ds \size {\cos M \int_0^A \dfrac {e^{-\alpha M} }{\alpha^2 + 1} \rd \alpha} + \size {\sin M \int_0^A \dfrac {\alpha e^{-\alpha M} }{\alpha^2 + 1} \rd \alpha }\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \int_0^A \dfrac {e^{-\alpha M} }{\alpha^2 + 1} \rd \alpha + \int_0^A \dfrac {\alpha e^{-\alpha M} }{\alpha^2 + 1} \rd \alpha\)
\(\ds \) \(\le\) \(\ds 2 \int_0^A e^{-\alpha M} \rd \alpha\)
\(\text {(3)}: \quad\) \(\ds \) \(\le\) \(\ds \dfrac 2 M\) similarly to $\paren 1$

Therefore:

\(\ds \size {\map {I_M} 0 - \dfrac \pi 2}\) \(=\) \(\ds \size {\paren {\map {I_M} A - \map {I_M} A } + \map {I_M} 0 + \paren {-\int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} + \int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} } -\dfrac \pi 2}\) adding zero
\(\ds \) \(=\) \(\ds \size {\map {I_M} A - \paren {\map {I_M} A -\map {I_M} 0 + \int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} } + \int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} -\dfrac \pi 2}\) rearranging
\(\ds \) \(\le\) \(\ds \size {\map {I_M} A} + \size {\map {I_M} A -\map {I_M} 0 + \int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} } + \size {\int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} - \dfrac \pi 2}\) Triangle Inequality for Real Numbers
\(\ds \) \(\le\) \(\ds \dfrac 1 A + \dfrac 2 M + \size {\int_0^A \dfrac {\rd \alpha} {\alpha^2 + 1} - \dfrac \pi 2}\) by $\paren 1$ and $\paren 3$
\(\ds \) \(\to\) \(\ds \dfrac 2 M\) as $A \to +\infty$ by Definite Integral to Infinity of $\dfrac 1 {x^2 + a^2}$

As:

$\ds \map {I_M} 0 = \int_0^M \dfrac {\sin x} x \rd x$

we have shown:

$\ds \forall M \in \R_{>0} : \size {\int_0^M \dfrac {\sin x} x \rd x - \dfrac \pi 2} \le \dfrac 2 M$

In particular:

\(\ds \int_0^\infty \dfrac {\sin x} x \rd x\) \(=\) \(\ds \lim_{M \mathop \to +\infty} \int_0^M \dfrac {\sin x} x \rd x\) Definition of Improper Integral
\(\ds \) \(=\) \(\ds \dfrac \pi 2\)

$\blacksquare$


Source of Name

This entry was named for Johann Peter Gustav Lejeune Dirichlet.