The integral of sin(x)/x dx from -infinity to infinity

Up a level : Integrals
Previous page : The integral of (e^(-px)-e^(-qx))/x dx from 0 to infinity
Next page : The integral of ln(x+1)/(x^2+1) dx from 0 to 1

We can use the result from the previous to solve a classical integral

$\int\limits_{ - \infty }^\infty {\frac{{\sin x}}{x}dx}$

We may first, for symmetry reasons, rewrite it as

$I = \int\limits_{ - \infty }^\infty {\frac{{\sin x}}{x}dx = 2} \int\limits_0^\infty {\frac{{\sin x}}{x}dx}$

There are several ways to continue from this, but here will use the result from the previous section – and a bit of complex analysis. We will use one of the formulas from the page on some trig identities. We have that

$sin(\theta ) = \frac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}}$

Substituting this into our integral we get

$I = 2\int\limits_0^\infty {\frac{{\sin x}}{x}dx} = 2\int\limits_0^\infty {\frac{{{e^{ix}} - {e^{ - ix}}}}{{2ix}}dx} = \frac{1}{i}\int\limits_0^\infty {\frac{{{e^{ix}} - {e^{ - ix}}}}{x}dx}$

But wait a minute, that’s just an integral of the form we solved in the previous section:

$\int\limits_0^\infty {\frac{{{e^{ - px}} - {e^{ - qx}}}}{x}dx} = - \ln (p) + ln(q) = ln\frac{q}{p}$

We thus have p= –i and q=i. This would give us

$\int\limits_0^\infty {\frac{{{e^{ix}} - {e^{ - ix}}}}{x}dx} = \ln \frac{i}{{ - i}} = \ln ( - 1)$

But since

$- 1 = {e^{ - i\pi }}$

we get

$\ln ( - 1) = \ln {e^{ - i\pi }} = - i\pi$

This would finally give us

$I = \int\limits_{ - \infty }^\infty {\frac{{\sin x}}{x}dx = } \frac{1}{i}i\pi = \pi$

One might wonder if one could actually do this, and if so, under what conditions – but the answer must wait until a future page.

Up a level : Integrals
Previous page : The integral of (e^(-px)-e^(-qx))/x dx from 0 to infinity
Next page : The integral of ln(x+1)/(x^2+1) dx from 0 to 1

Last modified: Sep 3, 2024 @ 15:35