Dirichlet Integral/Proof 5
Jump to navigation
Jump to search
Theorem
- $\ds \int_0^\infty \frac {\sin x} x \rd x = \frac \pi 2$
Proof
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac \pi 2\) |
$\blacksquare$