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So here is my question,

I would like to compute the following limit, $$\lim_{n\rightarrow\infty}\int_{\mathbb R_+}\frac{sin(x)}{x}(e^{-x/n}-1)dx$$ To interchange the integral an the limit I want to apply Dominated Convergence i.e i have to bound $|\frac{sin(x)}{x}(e^{-x/n}-1)|$ with some integrable function. I know I can obtain such a bound by integrating $\frac{sin(x)}{x}(e^{-x/n}-1)$ by parts. So I wanted to ask if someone knows a direct way to bound without using integration by parts? Thanks.

Edit

This is a part of an exercise where the goal is to prove that $\int_{\mathbb R_+}\frac{sin(x)}{x}=\pi/2$. So i am not allowed to use that $\int_{\mathbb R_+}\frac{sin(x)}{x}$ is finite.

Moreover in the solution of the exercise, where the one goal is to show that, $$\lim_{n\rightarrow\infty}|\int_{\mathbb R_+}\frac{sin(x)}{x}(e^{-x/n}-1)dx|=0$$ My professor wrote this to prove the above statement,

An integration by parts shows that, $$|\int_{\mathbb R_+}\frac{sin(x)}{x}(e^{-x/n}-1)dx|=|[\frac{1-cos(x)}{x}(e^{-x/n}-1)]_0^{\infty}-\int_{\mathbb R_+}\frac{1-cos(x)}{x^2}[1-e^{-x/n}(1+x/n)]dx|\leq|\int_{\mathbb R_+}\frac{1-cos(x)}{x^2}[1-e^{-x/n}(1+x/n)]dx|$$ Then he defines $g(x)_n:=\frac{1-cos(x)}{x^2}|1-e^{-x/n}(1+x/n)|$. Clearly $\lim_{n\rightarrow\infty}g(x)_n=0$. Furthermore since $g(x)_n\leq\frac{1-cos(x)}{x^2}$ and $\frac{1-cos(x)}{x^2}$ is integrable we can finally conclude, $$|\frac{sin(x)}{x}(e^{-x/n}-1)|\leq g(x)_n\rightarrow0$$ what proves the upper claim.

So i wanted to know if there is an easier way to obtain this?

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  • $\begingroup$ What will interchanging integral and limit do for you? (Sorry for this dumb question.) $\endgroup$
    – Jason Zimba
    Mar 25, 2014 at 13:02
  • $\begingroup$ Do you have $e^{-x/n}$ or $e^{-xn}$? $\endgroup$
    – user63181
    Mar 25, 2014 at 13:11
  • $\begingroup$ @SamiBenRomdhane we have $e^{-\frac{x}{n}}$ $\endgroup$
    – Bman72
    Mar 25, 2014 at 14:38

1 Answer 1

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You cannot bound the non-negative functions $$ f(x)=\frac{1-e^{-x/n}}{x}\cdot\left|\sin x\right| $$ or (it is clearly the same) $$ g_n(x) = \frac{1-e^{-x}}{x}\cdot\left|\sin(nx)\right| $$ with an integrable function over $\mathbb{R}^+$, since there exists a translation-invariant subset $H$ of $\mathbb{R}^+$ with infinite measure where $|\sin(n x)|\geq\frac{1}{2}$. Over $K=H\cap\{x\in\mathbb{R}:x>\log 2\}$ $$ g_{n}(x) > \frac{1}{4x} $$ holds, while $\frac{1}{4x}$ is not an integrable function over $K$.

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  • $\begingroup$ The main issue here is that $\frac{\sin x}{x}$ is Riemann-integrable over $\mathbb{R}$ but does not belong to $L^1(\mathbb{R})$! $\endgroup$
    – Jack D'Aurizio
    Mar 25, 2014 at 13:52
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    $\begingroup$ I just postet the solution of my professor to the problem, and in his solution the function must be integrable otherwise there is a mistake i am not able to find. I must also admit that I am not finding any mistake in your solution... Allready now thank you very much for helping me. $\endgroup$
    – Thorben
    Mar 25, 2014 at 14:03
  • $\begingroup$ The mistake is that the solution proves $|\int f|=|\int g|$ where $g$ is an $L^1$ function, and this is enough to state that $\frac{\sin x}{x}$ is Riemann-integrable. Clearly, the integral identity do not prove $|f|\leq |g|$, since the inequality cannot hold: $f\gg 1/x$, while $g=O(1/x^2)$. The dominated convergence theorem is used to prove that $g$ belongs to $L^1$, not $f$. $\endgroup$
    – Jack D'Aurizio
    Mar 25, 2014 at 15:40

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