6
$\begingroup$

in the evaluation of $\int_{0}^{+\infty}\frac{\sin x}{x}$, there was such a strategy as to evaluate it as $\lim_{s \to \infty}\int_{0}^{s}\frac{\sin x}{x}= \lim_{s \to \infty}\int_{0}^{s}\sin x \int_{0}^{+\infty}e^{-xt}dt $, and then we change the order of the integration by Fubini's theorem. but to use Fubini's theorem we have to know that the function is absolutely integrable in advance. How do we know that $\sin x e^{-xt}$ is absolutely integral in $(0,s)\times(0,+\infty)$?

$\endgroup$
  • 1
    $\begingroup$ The function $\frac{\sin x}{x}$ is not integrable on $(0,+\infty)$. But the improper integral exists and is $\frac{\pi}{2}$. See here. $\endgroup$ – Julien Apr 27 '13 at 14:04
  • 1
    $\begingroup$ This paper may be helpful too. $\endgroup$ – Raymond Manzoni Apr 27 '13 at 14:06
3
$\begingroup$

Use the fact that $|\sin x|\leq x$ for $x\geq 0$. The function $xe^{-xt}$ is integrable on $(0,s)\times (0,\infty)$ by Tonelli's theorem (this function is non-negative, and the iterated integral can be calculated explicitly if we integrate in the $t$ direction first). This means $\sin(x)e^{-xt}$ is integrable, since its absolute value is bounded by something integrable.

| cite | improve this answer | |
$\endgroup$

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.