Solving Integral of sinc(x) for Definite Integral

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SUMMARY

The definite integral of sinc(x), defined as \(\int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx\), equals \(\pi\). The discussion highlights various methods attempted to solve this integral, including power series expansions and trigonometric identities. Techniques such as double integration and polar coordinates, typically used for \(\int_{-\infty}^{\infty}e^{-x^2}dx\), were considered but deemed unhelpful. A notable approach involves using Euler's formula and integrating the expression \(\int_{0}^{\infty}\sin(x)\exp(-sx)dx=\frac{1}{1+s^2}\) from \(s = 0\) to infinity to derive the result.

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Homework Statement


I'm trying to prove the following definite integral of sinc(x)

\int_{-\infty}^{\infty}\frac{\sin(x)}{x}dx=\pi


Homework Equations





The Attempt at a Solution


I've tried power series expansions as well as trigonometric identities like

\frac{\cos 2x}{x}=\frac{\cos^2 x}{x}-\frac{\sin^2 x}{x}

I also looked at techniques used to integrate the definite integral
\int_{-\infty}^{\infty}e^{-x^2}dx

which I know is solved by double integration and changing to polar coordinates. However, this does not help me integrate sinc(x).
 
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Well, I suppose you could do it by making a closed curve in the complex plane and using Caychy's theorem (and Jordan's lemma). There might be an easier way, but I can't think of any.
 
Think about euler's formula and leibniz. A 'simple' proof can be made this way.
 
\int_{0}^{\infty}\sin(x)\exp(-sx)dx=\frac{1}{1+s^2}

Integrate both sides from s = 0 to infinity to obtain the result.
 
sinc(x) = sin(x)
x
has no anti-derivative
 
ahmedxahmed said:
sinc(x) = sin(x)
x
has no anti-derivative

no elementary anti-derivative :wink:
 

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