## Computing a definite Sine Integral

The definite Sine Integral

$\displaystyle\int_0^{\infty}\frac{\sin(x)}{x}dx = \frac{ \pi}{2}$

is often computed with contour integration. This requires knowledge of complex analysis. I just realized that it can also be computed very nicely with just real analysis. Instead of considering only this single integral define a function $f(t)$ depending on a non-negative real number $t$ by

${ f(t) = \displaystyle \int_0^{\infty} \frac{\sin(x) \, e^{-t x}}{x} dx}$.

Then $f(0)$ will be the integral that we want to compute. It is also clear that $\lim_{t \rightarrow \infty} f(t) = 0$, since the term $e^{-t x}$ becomes small very rapidly. Let’s first do some hand waving exploration. Assume that $f$ is differentiable and that its derivative can be computed by differentiating the integrand (as a function of $t$). Then we find

$f'(t) = -\displaystyle\int_0^{\infty} \sin(x) \, e^{-t x} dx$

This integral can be computed with an explicit anti-derivative for positive $t$.

$\dfrac{d}{dx} \left( \dfrac{(t \sin(x) + cos(x)) e^{-t x}}{t^2 + 1} \right) = -\sin(x) e^{-t x}$

Filling in the boundaries of integration we find

$f'(t) = -\dfrac{1}{t^2 + 1}$

And again, this equation can be solved explicitly: $f(t) = C - \arctan(t)$ for some constant $C$. This equation also holds for $t=0$ because $f$ was assumed to be differentiable and hence continuous. We also know that $f(t)$ tends to $0$ for $t \rightarrow \infty$ and that means that $C = \pi/2$ and in particular $f(0) = \pi/2$.

The hand waving at least results in the correct value for the definite Sine Integral! Instead of rigorously checking all steps above (which might be possible) I take an alternative approach and get rid of the infinity in the integration. For each positive integer $k$ define a function by

$f_k(t) = \displaystyle \int_0^{2 \pi k} \frac{ \sin(x) e^{-t x}}{x} dx$

Then $\lim_{t \rightarrow \infty} f_k(t) = 0$. Since the integrand is differentiable any number of times in both $x$ and $t$, there is now no problem in computing the derivative $f_k'(t)$ by differentiating the integrand

$f_k'(t) = -\displaystyle \int_0^{2 \pi k} \sin(x) e^{-t x} dx = \frac{e^{-2 \pi k t} - 1}{t^2 +1}$

Integrating this equation and plugging in the correct value for $f_k(0)$ we find

$f_k(t) = \displaystyle \int_0^{2 \pi k} \frac{\sin(x)}{x} dx - \arctan(t) + \int_0^t \frac{e^{-2 \pi k s}}{s^2 + 1}ds$

Taking the limit for $t \rightarrow \infty$ this results in

$\displaystyle \int_0^{2 \pi k} \frac{ \sin(x) }{x}dx = \frac{ \pi }{2} - \int_0^{\infty} \frac{e^{-2 \pi k s}}{s^2 + 1}ds$

The last integral term in this equality tends to zero as $k \rightarrow \infty$ and taking this limit for $k$ finally result in the value $\pi/2$ for the definite Sine Integral.

Anyone familiar with the Laplace transform would probably summarize this post as a one liner

${\cal L}\{ \dfrac{ \sin(x) }{x} \} = \dfrac{ \pi }{2} - \arctan(t) = \arctan(\dfrac{1}{t})$