The integral of the Sinc function over $\mathbb{R}$ is well-known, $$\int_{-\infty}^{\infty} \frac{\sin(x)}{x} dx = \pi$$But if I try to evaluate this using integration by parts, I get $$\int_{-\infty}^{\infty} \frac{\sin(x)}{x} dx = \left.-\frac{\cos(x)}{x} \right|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2} dx$$ The first part is $0$, and the second part diverges.

What's going on? Is integration by parts just not kosher here? If so, why not?


The definite integration by parts formula

$$\int_a^b f(x) g(x)\ dx = F(x) g(x)\bigg|_a^b - \int_a^b F(x) g'(x)\ dx$$

is justified by the product rule for derivative and the fundamental theorem of calculus

$$\int_a^b (F g)'(x)\ dx = F(x) g(x) \bigg|_a^b $$

But the "fine print" there is that $F(x) g(x)$ is assumed to be continuously differentiable on the interval $(a,b)$ and continuous on $[a,b]$. In this case with $F(x) = -\cos(x)$ and $g(x) = 1/x$, $F(x) g(x)$ is undefined at $x=0$, and no definition of it there will make it continuous at $x=0$, let alone differentiable. So you can't use the integration by parts formula for this integral on any interval containing $0$.


The issue is the discontinuity at $0$ in the cosine terms when you perform the partial integration, which is overlooked if you integrate over $\mathbb{R}$. To get around this, consider:

$$\int_{-\infty}^{\infty}\frac{\sin x}{x}dx=2\int_0^{\infty}\frac{\sin x}{x}dx$$

By parts:

$$\int_{0}^{\infty} \frac{\sin(x)}{x} dx = \left.-\frac{\cos(x)}{x} \right|_{0}^{\infty} - \int_{0}^{\infty} \frac{\cos(x)}{x^2} dx$$

And here, both terms on the RHS diverge. No contradiction.

  • $\begingroup$ There is a discontinuity in $\cos(x)/x$, though not in $\sin(x)/x$. $\endgroup$ – Robert Israel Jan 28 '13 at 6:26
  • $\begingroup$ @RobertIsrael I thought the OP meant a discontinuity at zero of the function $\sin x/x$. $\endgroup$ – user38268 Jan 28 '13 at 6:27
  • $\begingroup$ @BenjaLim Sorry I realise that wasn't absolutely clear in my original post. $\endgroup$ – L. F. Jan 28 '13 at 6:45
  • $\begingroup$ @L.F., this proof is not accurate. You can not integrate by parts since you have an integrand that is not convergent. If the denominator was $x^{-2}$, than okay, but this is not the case. $\endgroup$ – Jeff Faraci Dec 15 '13 at 3:44
  • $\begingroup$ @Jeff I haven't proved anything $-$ please re-read the question. $\endgroup$ – L. F. Dec 15 '13 at 3:48

As others have already discussed, integrating by parts with $u=\frac1x$ and $v=-\cos(x)$ fails due to the singularity at $0$. Here, I thought it would be instructive to present an integration by parts scheme that circumvents the difficulty of the singularity at $0$. To that end, we proceed.

Let $I$ be the integral given by

$$I=\int_0^\infty \frac{\sin(x)}{x}\,dx\tag1$$

Enforcing the substitution with $u=\frac1x$ and $v=1-\cos(x)$ (instead of $\displaystyle v=-\cos(x)$) in $(1)$ yields

$$\begin{align} I&=\left.\left(\frac{1-\cos(x)}{x}\right)\right|_0^\infty+\int_0^\infty \frac{1-\cos(x)}{x^2}\,dx\\\\\ &=2\int_0^\infty \left(\frac{\sin(x/2)}{x}\right)^2\,dx\\\\ &=\int_0^\infty \left(\frac{\sin(x)}{x}\right)^2\,dx\tag2 \end{align}$$

While $(2)$ does not lead immediately to a value for $I$ it does reveal the interesting identity

$$\int_0^\infty\frac{\sin(x)}{x}\,dx=\int_0^\infty \left(\frac{\sin(x)}{x}\right)^2\,dx$$


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