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

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?


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 $t > 0$.

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

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