Evaluate the Integral $\int_{-\infty}^{+\infty}\text{sinc}(x)dx$.

Question

Evaluate the integral $$ \int_{-\infty}^{+\infty}\text{sinc}(x)\text{ d}x. $$ My Try : let $$ I=\int_{-\infty}^{+\infty}\text{sinc}(x)\text{ d}x. $$ Now by definition, we have $I=\int_{-\infty}^{+\infty}\frac{\sin x}{x}\text{ d}x$. Note that the function is even, so we can write $$ I=2\int_{0}^{\infty}\frac{\sin x}{x}\text{ d}x. $$ Now let's define for $a>0$, $$ K(a)=\int_{0}^{\infty}\frac{e^{-ax}\sin x}{x}\text{ d}x. $$ From here we have $$ I=2\lim_{a\to 0}K(a) $$ and also observe that $K(a)$ is continuous at $a=0$. There I got stuck. Can anyone suggest me of how can I solve the rest? Also is there any alternate method to solve this problem? Any help would be appreciated .

Answer 1

Your approach is good. Consider expanding $\sin x$ in a Maclaurin series, you should find that: $$\int_0^\infty\frac{1}{x}\sin(x)e^{-ax}\,\mathrm{d}x=\arctan(1/a)$$

You might want to be a little careful about justifying continuity at zero. Notice the integral does not exist for negative $a$. Notice we can write: $$0\le\int_0^\infty\frac{\sin x}{x}(1-e^{-ax})\,\mathrm{d}x\le a\int_0^\pi\sin(x)\,\mathrm{d}x+\int_\pi^\infty\frac{\sin x}{x}(1-e^{-ax})\,\mathrm{d}x$$The first of the latter integrals vanishes as $a\to0^+$ of course. The second can be shown to vanish by carefully breaking the domain of integration into pieces $[\pi n,\pi(n+1)]$ and making suitable bounds.

Answer 2

Here is my non-rigorous approach:

\begin{align*} \Omega &:= \int_{-\infty}^\infty \mathrm{sinc}(x)\,dx = 2 \int_0^\infty \frac{\sin(x)}{x}\,dx. \end{align*}

We now define a function in terms of $t$: \begin{align*} f(t) := \int_0^\infty \frac{\sin(tx)}{x}\,dx. \end{align*}

Notice that $2f(1) = \Omega$. We apply a the Laplace transform to our function: \begin{align*} \mathcal{L}_t\{f(t)\} &= \mathcal{L}\left\{\int_0^\infty \frac{\sin(tx)}{x}\,dx\right\}\\ &= \int_0^\infty e^{-st} \int_0^\infty \frac{\sin(tx)}{x}\,dx \,dt\\ &= \int_0^\infty \frac{1}{x} \int_0^\infty e^{-st}\sin(tx)\,dt\,dx\\ &= \int_0^\infty \frac{1}{x} \mathcal{L}_t\{\sin(tx)\}\,dx\\ &= \int_0^\infty \frac{1}{x} \cdot \frac{x}{s^2 + x^2}\,dx\\ &= \int_0^\infty \frac{1}{s^2 + x^2}\,dx\\ &= \lim_{R \to \infty} \left[\frac{1}{s} \arctan\left(\frac{x}{s}\right) \right]_{x =0}^R\\ &= \frac{\pi}{2s}. \end{align*}

To get our closed form for $f(t)$, we now apply the Inverse Laplace transform to $\mathcal{L}_t\{f(t)\}$: \begin{align*} f(t) &= \mathcal{L}^{-1}\{\mathcal{L}_t\{f(t)\}\}\\ &= \mathcal{L}^{-1}\left\{\frac{\pi}{2s}\right\}\\ &= \frac{\pi}{2}. \end{align*}

Now that we know that $f(t) = \frac{\pi}{2}$, we know that $f(1) = \frac{\pi}{2}$. Thus, we can conclude that \begin{align*} \Omega &= 2 f(1) = 2 \cdot \frac{\pi}{2} = \pi. \end{align*}

Answer 3

Here is another quick approach for Fourier transform lovers.

Since the Fourier transform of $\mathrm{sinc}$ is $$ {\cal F}(\mathrm{sinc})(y) = \int_{-\infty}^{\infty} \frac{\sin(x)}{x} \,e^{-2i\pi \,x\cdot y}\,\mathrm d y = \pi\,{\bf 1}_{[-1/\pi,1/\pi]}(y) $$ as can be obtained from the Fourier inversion theorem, one deduces that $$ \int_{-\infty}^{\infty} \frac{\sin(x)}{x} \,\mathrm d y = {\cal F}(\mathrm{sinc})(0) = \pi. $$

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Remark: The $L^2$ definition of the Fourier transform is sufficient here, as $\mathrm{sinc}$ is square integrable. This implies that the integral can be understood as the limit of the integral on $[-n,n]$ for $n\to\infty$.