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

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

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