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})

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