# $\int_{0}^{\infty}\frac{\sin(x)}{x}dx$ exists, but $\int_{\mathbb{R^+}}\frac{\sin(x)}{x}d\lambda$ doesn't exist?

$$\int_{0}^{\infty}\frac{\sin(x)}{x}dx$$ is riemann integrable.

But how to prove that it's not Lebesgue integrable ?

If $$f(x)=\frac{\sin(x)}{x}$$, I supppose that $$f\in L^1(\mathbb{R^+})$$.

then, I' ll define the function sequence : $$f_n(x)=\frac{\sin( x)}{x}\mathbb{1}_{[0,n]}(x).$$

We have $$f_n$$ converges $$\lambda$$-a.e to $$f$$.

and $$|f_n|\leq|f|$$, for all $$1\leq n$$.

and $$f$$ integrable (hypothesis).

then with the dominated convergence theorem

$$\lim_n\int_{0}^{n}f_nd\lambda=\int_{\mathbb{R^+}}fd\lambda$$

which is equivalent to $$\int_{0}^{\infty}\frac{\sin(x)}{x}dx=\int_{\mathbb{R^+}}f d\lambda$$

Thats basically what I did.

• Hi Anas, as part of the community guidelines you need to provide some more context and background to your question. Furthermore if you have attempted the problem you need to provide your working. Just a heads up :-) – user150203 Dec 23 '18 at 7:19
• @DavidG, check . – Anas BOUALII Dec 23 '18 at 7:31
• Looks good to me! Just wanted to let you know. This page has very specific rules about posting. It annoyed me to start with, but once you get use to it you realise it's a great format to work with. Sorry I can't help you wing your question though. – user150203 Dec 23 '18 at 7:35
• Well, i can't find the contradiction with the last equality. but thanks anyway. – Anas BOUALII Dec 23 '18 at 7:38
• Actually it is not Riemann integrable either. It is only improperly Riemann integrable. – metamorphy Dec 23 '18 at 7:43

Hint. If such function is Lebesgue integrable then also its absolute value is Lebesgue integrable. Let $$n$$ a positive integer, then $$\int_0^{2\pi n} \frac{|\sin x|}{x} \,dx = \sum_{k=0}^{n-1} \int_{2\pi k}^{2\pi(k+1)} \frac{|\sin x|}{x}\,dx \ge \sum_{k=0}^{n-1} \frac{1}{2\pi (k+1)} \int_{0}^{2\pi} |\sin x|\,dx$$ Can you take it from here?