A type of integral that's generalized by [[Dirichlet's principle]], the Dirichlet integral is defined as
$I =\int_0^\infty dx \frac{\sin(x)}{x} = \frac{\pi}{2}$
where the integrand has to be a [Continuous function](Continuous%20function.md) on every interval $[0,\gamma]$ for $\gamma \rightarrow \infty$.
Notice that since this is an even function,
$I =\int_{-\infty}^\infty dx \frac{\sin(x)}{x} = \pi.$
A notable fact about this integral is that it is not possible to evaluate as a [[Lebesgue integral]], however it is nonetheless an improper [[Riemann integral]].
# Solution as a [[Real integral on the complex plane]]
We start by taking $I$ to be the imaginary component of some complex number:
$I = \mbox{Im}(I')=\mbox{Im}\bigg(\int_{-\infty}^{\infty}dz \frac{e^{iz}}{z}\bigg)$
We should immediately notice that the complex integrand has a [[pole]] at $z=0$, which is on the real axis, part of the contour we need to integrate along. Hence we need to evaluate $I'$ by finding the [[Cauchy principal value]]. This is done by splitting the integral into two parts that are split by the origin that together will correspond to the cauchy principle value:
$I'=\lim_{\epsilon \rightarrow 0}\bigg[\int_{-\infty}^{-\epsilon}dz \frac{e^{iz}}{z} + \int_{\epsilon}^{\infty}dz \frac{e^{iz}}{z}\bigg] = \mbox{P.V.}\int_{-\infty}^{\infty}dz \frac{e^{iz}}{z}$
This procedure removes an infinitessimally small point at the origin where the singularity lies. Thus, in defining the integration contour we also define an infinitessimal semicircle over the origin and the contour integral along the length of the real axis excluding the origin is the Cauchy Principle Value. This splits the contour integral into segments $\Gamma_1$, $\Gamma_2$, $\Gamma_3$ as shown below:
Image adapted from
[Murayama H., MH2801: Complex Methods for the Sciences](08_contour_integration.pdf)
Here the integral over the arc $\Gamma_1$ is the [[Cauchy principle value]], $I'$.
The entire closed contour is, $\int_{\Gamma_1+\Gamma_2+\Gamma_3}dz \frac{e^{iz}}{z}=\oint_\Gamma dz \frac{e^{iz}}{z} = 0$
by [[Cauchy's theorem]]. And thus thus our next step is to solve for $I'$. where,
$0 = \int_{\Gamma_1}dz \frac{e^{iz}}{z}+\int_{\Gamma_2}dz \frac{e^{iz}}{z}+\int_{\Gamma_3}dz \frac{e^{iz}}{z} = I'+\int_{\Gamma_2}dz \frac{e^{iz}}{z}+\int_{\Gamma_3}dz \frac{e^{iz}}{z}$
The integral over the semicircular arc, $\Gamma_2$ is:
$\lim_{R\rightarrow \infty}\int_{\Gamma_2}dz \frac{e^{iz}}{z} = \lim_{R\rightarrow \infty} iR \int_{0}^{\pi} d\phi \; i\frac{R}{R}e^{i\phi}e^{e^{i\phi}}e^{-i\phi} = \lim_{R\rightarrow \infty} \int_{0}^{\pi} d\phi \;ie^{Re^{i\phi}}=0$
where we _did not_ need to actually evaluate this integral directly and instead note that since the integrand is analytic at all points above $R_0=0$ [condition 1 of Jordan's Lemma](Jordan's%20lemma.md) is met. Condition 2 is met by the form of the contour, $\Gamma_2$. And condition 3. of the lemma is met by showing that $|f(z)| = 1/|z| = 1/R^2$ and $| 1/R^2| \rightarrow 0$ as $R\rightarrow \infty$.
Integrating over $\Gamma_3$ can be done directly by [parameterizing](Closed%20contour%20integral.md#Direct%20calculation%20by%20parameterizing) with $R=\epsilon$ or by [using the residue theorem.](Dirichlet%20integral.md#Using%20the%20Residue%20theorem)
If we solve by parameterizing we find that
$\int_{\Gamma_3}dz\frac{e^{iz}}{z}=\lim_{\epsilon \rightarrow 0}\int_{\pi}^{0} d\phi \; i\epsilon e^{i\phi}\frac{e^{i\epsilon e^{i\phi}}}{\epsilon e^{i\phi}}= -i \pi$
The negative sign comes about as a result of integrating over a _negatively oriented_ or _clockwise_ contour.
Putting everything together and solving for $I'$ we find that $I'=i\pi$, and thus
$I=\mbox{Im}(i\pi)=\pi$
## Using the Residue theorem
The next to last step in the [above calculation](Dirichlet%20integral.md#Solution%20as%20a%20real%20integral%20on%20the%20complex%20plane) may be simplified with the [the residue theorem](Residue%20theorem.md#The%20Residue%20Theorem%20for%20infinitessimal%20circular%20arcs) applied on a negatively (clockwise) oriented infinitessimal half-circle as shown below.
Notice that $\mathrm{Res}_{z=0}\frac{e^{iz}}{z}=1.$ And along an infinitessimal half circle the residue theorem is expressed as
Thus the integral along $\Gamma_3$ is evaluated as
$\int_{\Gamma_3}dz\frac{e^{iz}}{z}= \pi i \sum_i \mbox{Res}_{z\rightarrow z_0}\Big(\frac{e^{iz}}{z}\Big)=-i\pi$
where the minus sign is accounted for by the fact that the contour is negatively oriented and $\alpha =\pi$ since the contour is a semicircle.
This is a valid application of the residue theorem since $\Gamma_3$ is not in fact an open contour but an infinitessimal semicircle.
# Solution using the Dominated Convergence Theorem
[Lebesgue's dominated convergence theorem](Lebesgue's%20dominated%20convergence%20theorem.md)
# Solution using a [[Laplace transform]]
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# Recommended reading
A step by step solution on the complex plane along with additional context is provided by the following set of lecture notes:
* [Murayama H., MH2801: Complex Methods for the Sciences](08_contour_integration.pdf)
For a solution using the Dominated Convergence Theorem see:
* [Bartle R., Sherbert D., _Introduction to Real Analysis_](Robert%20G.%20Bartle,%20Donald%20R.%20Sherbert%20-%20Introduction%20to%20Real%20Analysis%20%20Fourth%20Edition%20(2011,%20John%20Wiley%20&%20Sons).pdf) pgs. 319 and 320
#MathematicalFoundations/Analysis/Integrals
#MathematicalFoundations/Analysis/ComplexAnalysis/Integrals