A way to prove $\int_0^\infty \sin(x) / x dx = \frac{\pi}{2}$ is to switch integrals as in

$$\tag{1}\int_0^\infty \int_0^\infty e^{-xy} \sin(x) dx dy= \int_0^\infty \int_0^\infty e^{-xy} \sin(x) dy dx $$

because $\int_0^\infty e^{-xy} \sin(x) dy = \sin(x)/x$ and twice integration by parts gives $\int e^{-xy}\sin(x) dx = C - e^{-xy}(y \sin(x) + \cos(x))/(1 + y^2)$, etc.

My question here is how to justify switching the integrals in equation 1 without computing?

Fubini theorem is not helpful since $e^{-xy}\sin(x)$ is not nonnegative and not absolutely integrable on $\mathbb{R}^2$.

  • $\begingroup$ But we are on $[0,\infty)^2$ and not on $\mathbb R^2$. $\endgroup$ – zoli Dec 10 '17 at 23:51
  • $\begingroup$ That is what I meant $\endgroup$ – WoodWorker Dec 10 '17 at 23:52

The integrand is definitely not absolutely integrable. You can justify changing the order of integration by a combination of uniform and dominated convergence.

Note that by the Weierstrass test, the improper integral

$$\tag{1} F(x) = \int_0^\infty e^{-xy} \sin x \, dy$$

is uniformly convergent for $x$ in any compact interval.

Take sequences $a_n,c_m \to 0$ and $b_n,d_m \to +\infty$. Since the integrand is continuous the interchange is permitted on the bounded rectangle:

$$\int_{a_n}^{b_n} \int_{c_m}^{d_m}e^{-xy} \sin x \, dx \, dy= \int_{c_m}^{d_m} \int_{a_n}^{b_n}e^{-xy} \sin x \, dy \, dx.$$

By the uniform convergence of the improper integral (1), it follows that $$\tag{2}\int_{0}^{\infty} \int_{c_m}^{d_m}e^{-xy} \sin x \, dx \, dy= \lim_{n \to \infty}\int_{c_m}^{d_m} \int_{a_n}^{b_n}e^{-xy} \sin x \, dy \, dx \\ = \int_{c_m}^{d_m} \lim_{n \to \infty}\int_{a_n}^{b_n}e^{-xy} \sin x \, dy \, dx \\ = \int_{c_m}^{d_m} \int_{0}^{\infty}e^{-xy} \sin x \, dy \, dx. $$

If we can show that the limit of the LHS of (2) as $m \to \infty$ can be passed under the integral then we are done.

Note that

$$\left|\int_{c_m}^{d_m} e^{-xy} \sin x \, dx \right| = \left| \left.\frac{-e^{-xy} (y\sin x + \cos x)}{1+y^2}\right|_{x= c_m}^{x = d_m} \right| \leqslant \frac{C}{1+y^2}.$$

This bound holds because

$$\left| \frac{e^{-xy} \cos x}{1+y^2}\right| = \frac{e^{-xy}|\cos x|}{1 +y^2} \leqslant \frac{1}{1+y^2},$$

and $f(x) = ye^{-xy} \sin x $ has a maximum for $x \in [0,\infty)$ at $x^{*} = \arctan(1/y)$ where

$$f(x^{*}) = y e^{-y \arctan(1/y)} \sin \arctan(1/y) = ye^{-y \arctan(1/y)}\frac{1/y}{\sqrt{1 + (1/y)^2}} \leqslant 1 $$

Since $1/(1+y^2)$ is integrable we can apply DCT in taking the limit of both sides of (2) to obtain the desired result.

  • $\begingroup$ Question: how do you get that final inequality? Should not there be a y in the numerator? In that case it would not be integrable $\endgroup$ – WoodWorker Dec 11 '17 at 0:07
  • $\begingroup$ Do you mean the $C/(1 + y^2)$? $\endgroup$ – RRL Dec 11 '17 at 0:08
  • $\begingroup$ yes, that is what I mean $\endgroup$ – WoodWorker Dec 11 '17 at 0:09
  • $\begingroup$ The part with $\cos x$ term is no problem since $|e^{-xy} \cos x| /(1 +y^2) \leqlslant 1/(1+y^2)$ for all $x$. The other part is trickier but it can be shown that $|e^{-xy} y \sin x|$ is also bounded by a constant. $\endgroup$ – RRL Dec 11 '17 at 0:14
  • $\begingroup$ @WoodWorker: I added an explanation for the upper bound. $\endgroup$ – RRL Dec 11 '17 at 0:35

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