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.