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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!

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  • $\begingroup$ 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? $\endgroup$ – Michael Maier Jan 11 at 12:14
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$\frac {\sin\, x} x$ 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 $x \to \infty$].

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  • $\begingroup$ 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. $\endgroup$ – Michael Maier Jan 11 at 12:37
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    $\begingroup$ Surely if you're doing Lebesgue measure, you know how to evaluate a definite integral. Are you required to use only things covered in the course? $\endgroup$ – Riccardo Orlando Jan 11 at 12:38
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    $\begingroup$ @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. $\endgroup$ – Kavi Rama Murthy Jan 11 at 12:45
  • $\begingroup$ 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. $\endgroup$ – Michael Maier Jan 11 at 12:48
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    $\begingroup$ @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 $e^{-\alpha x} \leq e^{-\alpha a_n} =\frac 1 {n^{2}}$). $\endgroup$ – Kavi Rama Murthy Jan 11 at 12:49

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