# Show that $f$ is integrable with respect to the Lebesgue measure

Let $$\alpha \gt 0 \forall \alpha \in\mathbb{R}$$, $$f_{\alpha}:]0,\infty[\to\mathbb{R}$$, $$x\mapsto e^{-\alpha x}\left(\frac{sin(x)}{x}\right)^3$$. The goal ist to show that $$f_{\alpha}$$ is integrable with respect to the Lebesgue-measure, which coincides with the Borel measure in this case. I already showed that $$f_{\alpha}$$ is measurable and know that the condition for $$f_{\alpha}$$ being integrable is that $$\int_X\lvert{f(x)}\rvert\mathrm{d}\lambda\lt\infty$$ with $$X=]0,\infty[$$. Since I am fairly new to this topic, I don't know where to start. I struggle to write out the integral since it can attain negative values and lack a general strategy for exercises like this. Any help is greatly appreciated!

• I guess it would be helpful to split the integral in two parts and then write it down explicitly and then to hope that the sum of my $\lambda$-measures somehow converges? – Michael Maier Jan 11 at 12:14

$$\frac {\sin\, x} x$$ is abounded function on $$(0,\infty)$$. If $$|\frac {\sin\, x} x| \leq M$$ then $$\int_0^{\infty} |f_{\alpha} (x)| \, dx \leq M^{3}\int_0^{\infty} e^{-\alpha x}\, dx =\frac {M^{3}} {\alpha}$$. [ Boundedness of $$\frac {\sin\, x} x$$ is a consequence of the following facts: this function is continuous, it approaches $$1$$ as $$x\to 0$$ and approaches $$0$$ as $$x \to \infty$$].
• Thanks. Problem is, we didn't cover $\int_0^\infty e^{-\alpha x}dx=\frac{1}{\alpha}$. I just need to show that $\int_0^\infty e^{-\alpha x}d\lambda$ is integrable then, any advice on that? Obviously it is decreasing, but I don't see where that might help. – Michael Maier Jan 11 at 12:37
• @MichaelMaier Riemann integrable functions are Lebesgue integrable and their Lebesgue integral is same as Riemann integral. I am sure you have covered intergral of $e^{-\alpha x}$ before starting with measure theory. – Kavi Rama Murthy Jan 11 at 12:45
• Thanks again. As a physics student, I indeed know what the integral of $e^{-\alpha x}$ is. Problem is, our math lectures are unrelated to that and we do just mention Riemann integrals later on. So I do know how to calculate the integral, but am not allowed to use anything related to Riemannian Integration in my proof. – Michael Maier Jan 11 at 12:48
• @MichaelMaier If $a_n=\frac 2 {\alpha} \log\, n$ then $a_n$ increases to $\infty$. Split the integral of $e^{-\alpha x}$ into integrals over the intervals $(a_n,a_{n+1})$. Can use this to show that the integral is finite? (On $(a_n,a_{n+1})$ note that $e^{-\alpha x} \leq e^{-\alpha a_n} =\frac 1 {n^{2}}$). – Kavi Rama Murthy Jan 11 at 12:49