Dirichlet Integral/Proof 2
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Theorem
- $\ds \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
Proof
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$\ds \int_0^\infty \frac {\sin x} x \rd x$ 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:
- $\ds \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
$\blacksquare$