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How can I evaluate the following improper integral:

$$ \int\limits_0^{\infty}\frac{\sin x}{x}\, dx$$

I have tried to evaluate this by integration by parts but failed. Please help

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marked as duplicate by user99914, астон вілла олоф мэллбэрг, user186170, Sangchul Lee, Winther Sep 15 '16 at 12:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ @JohnMa thanks, this link is helpful. Can we evaluate this integral by using the properties of beta function or gamma function? $\endgroup$ – user356595 Sep 15 '16 at 10:02
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The integral is convergent by Dirichlet's test. We may notice that $$ I=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\sin x}{x}\,dx = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\sin(x)\cdot\color{green}{\left[\frac{1}{x}+\sum_{n\geq 1}(-1)^n\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right)\right]}dx\tag{1}$$ hence it is enough to understand what kind of function is the green function.
By the Weierstrass product for the sine function: $$ \sin(x) = x\cdot\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)\tag{2}$$ and by considering the logarithmic derivative of both sides: $$ \cot(x) = \frac{1}{x}+\sum_{n\geq 1}\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right) \tag{3} $$ that leads to: $$ \frac{1}{\sin(x)}=\frac{1}{x}+\sum_{n\geq 1}(-1)^n\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right)\tag{4} $$ that is exactly the green function. It follows that: $$ I = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{\sin x}{\sin x}\,dx = \color{red}{\frac{\pi}{2}}.\tag{5} $$

This is an explanation of qoqosz' answer here.

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you define the function $$F(t)=\int_0^\infty \frac{sinx}{x}e^{-xt}dx$$ you show that $F$ satisfies a simple differential equation. you find $F(t)$ and $F(0)$ gives you the result $\frac{\pi}{2}$.

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