2
$\begingroup$

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 $\mathbb{R}^2$.

$\endgroup$
  • $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.