# Linked Questions

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### Proving $\int\limits _0^\infty\frac {\sin^{2n+1}x}{x}dx=\frac 1 {4^n}\binom{2n}{n}\int_0^\infty \frac {\sin{x}}{x}dx$

The question is Prove that for any $n\in\mathbb N$, $\int\limits _0^\infty\frac {\sin^{2n+1}x}{x}dx=\frac 1 {4^n}\binom{2n}{n}\int_0^\infty \frac {\sin{x}}{x}dx$ I don't have any ideas how to ...
2answers
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### Parameterizing divots in contour integrals

I have recently attempted to pick up complex analysis and have been stuck on this problem for a few days: $$\int_{-\infty}^\infty \frac{\sin^2x}{x^2}\mathrm{d}x$$ Fortunately, it would seem that ...
0answers
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1answer
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I've already proved that the improper integral $\int_{0}^{\infty}\frac{\operatorname{sin}(t)}{t}$ is convergent. I don't know its limit though... I'm asked to prove that $\begin{array}{ccccc} f ... 1answer 38 views ### convergence dominate of Lebesgue-Application my question is: let$f(x)= \dfrac{\sin (nx)}{x}.\varphi(x)$, where$\varphi \in C^\infty_c(\mathbb{R})$. We can found an function$g \in L^1(\mathbb{R})$such as$|f(x)| \leq g$? Please. I think that, ... 2answers 49 views ### How to study the convergence of$\int^{\infty}_{0}\frac{2\cos(t)\sin(t)}{t}dt$? [duplicate] By using Taylor series, I managed to see that the$t \rightarrow 0$,$\frac{2\cos(t)\sin(t)}{t} = 1 -\frac{4t^2}{6}$whose integral converges. As for the case when$t \rightarrow \infty\$, I don't know ...

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