1
$\begingroup$

$$f(t)=\int_{0}^{+\infty}e^{-tx}\frac{\sin x}{x}dx, t\in\mathbb{R}$$

I need to find out for which $t \in \mathbb{R}$ this integrals exists (meaning it doesn't diverge) as Riemann-integral at first and then as Lebesgue-integral.

As a Riemann integral it exists for $t \ge0$. But how can I show this or how can I show that it doesn't exist $t <0$?

And what about the Lebesgue integral?

$\endgroup$
0
$\begingroup$

The Cauchy criterion can be helpful. The improper integral $\displaystyle \int_0^\infty f(x) \, dx$ exists and is finite if and only if $$\displaystyle \lim_{n,m \to \infty} \int_n^m f(x) \, dx = 0.$$

Let $t < 0$. Integrate over a half-period of the $\sin$ function to estimate $$\int_{2k\pi}^{2k\pi + \pi} e^{-tx} \frac{\sin x}{x} \, dx = \int_{2k\pi}^{2k\pi + \pi} e^{|t|x} \frac{\sin x}{x} \, dx \ge \frac{e^{2k\pi|t|}}{2k\pi+\pi}\int_{2k\pi}^{2k\pi + \pi} \sin x\, dx= \frac{2e^{2k\pi|t|}}{2k\pi+\pi} \to \infty$$ as $k \to \infty$. The Cauchy property fails, so the improper integral diverges.

Just about the same argument shows you that if $t \le 0$ then $$\int_{0}^{\infty} \left| e^{-tx} \frac{\sin x}{x} \right| \, dx = \infty$$ so the Lebesgue integral does not exist for all $t \le 0$.


If $t > 0$ use the fact that $\displaystyle \left| e^{-xt} \frac{\sin x}{x} \right| \le e^{-xt}$ and $\displaystyle \int_0^\infty e^{-xt} \, dt < \infty$ to conclude the Lebesgue integral exists for $t > 0$.

$\endgroup$
  • $\begingroup$ And why does the Lebesgue integral exist for $t>0$? $\endgroup$ – conrad Nov 3 '18 at 9:56

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.