Editing Dirichlet integral (section)

=== Differentiation under the integral sign (Feynman's trick)===
First rewrite the integral as a function of the additional variable <math>s,</math> namely, the Laplace transform of <math>\frac{\sin t} t.</math> So let
<math display="block">f(s)=\int_0^\infty e^{-st} \frac{\sin t} t \, dt.</math>

In order to evaluate the Dirichlet integral, we need to determine <math>f(0).</math> The continuity of <math>f</math> can be justified by applying the [[dominated convergence theorem]] after integration by parts. Differentiate with respect to <math>s>0</math> and apply the [[Leibniz integral rule|Leibniz rule for differentiating under the integral sign]] to obtain
<math display="block">
\begin{align}
\frac{df}{ds} & = \frac{d}{ds}\int_0^\infty e^{-st} \frac{\sin t}{t} \, dt = \int_0^\infty \frac{\partial}{\partial s}e^{-st}\frac{\sin t} t \, dt \\[6pt]
& = -\int_0^\infty e^{-st} \sin t \, dt.
\end{align}
</math>

Now, using Euler's formula <math>e^{it} = \cos t + i\sin t,</math> one can express the sine function in terms of complex exponentials:
<math display="block">
\sin t = \frac{1}{2i} \left( e^{i t} - e^{-it}\right).
</math>

Therefore,
<math display="block">
\begin{align}
\frac{df}{ds} & = -\int_0^\infty e^{-st} \sin t \, dt = -\int_{0}^{\infty} e^{-st} \frac{e^{it} - e^{-it}}{2i} dt \\[6pt]
&= -\frac{1}{2i} \int_{0}^{\infty} \left[ e^{-t(s-i)} - e^{-t(s + i)} \right] dt \\[6pt]
&= -\frac{1}{2i} \left [ \frac{-1}{s - i} e^{-t (s - i)} - \frac{-1}{s + i} e^{-t (s + i)}\right]_0^{\infty} \\[6pt]
&= -\frac{1}{2i} \left[ 0 - \left( \frac{-1}{s - i} + \frac{1}{s + i} \right) \right] = -\frac{1}{2i} \left( \frac{1}{s - i} - \frac{1}{s + i} \right) \\[6pt]
&= -\frac{1}{2i} \left( \frac{s + i - (s -i)}{s^2 + 1} \right) = -\frac{1}{s^2 + 1}.
\end{align}
</math>

Integrating with respect to <math>s</math> gives
<math display="block">f(s) = \int \frac{-ds}{s^2 + 1} = A - \arctan s,</math>

where <math>A</math> is a constant of integration to be determined. Since <math>\lim_{s \to \infty} f(s) = 0,</math> <math>A = \lim_{s \to \infty} \arctan s = \frac{\pi}{2},</math> using the principal value. This means that for <math>s > 0</math> 
<math display="block">f(s) = \frac{\pi}{2} - \arctan s.</math>

Finally, by continuity at <math>s = 0,</math> we have <math>f(0) = \frac{\pi}{2} - \arctan(0) = \frac{\pi}{2},</math> as before.