Find $\int_{-\infty}^{\infty}\frac{sin^2(t)}{t^2}dt$

Asked Apr 29 '21 at 00:58

Active Apr 29 '21 at 00:58

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Let $$f(z) = \frac{1-e^{2iz}}{z^2}$$

and let R > r > $0$

Find $$\int_{-\infty}^{\infty}\frac{sin^2(t)}{t^2}dt$$

using

$$\int_{\gamma}f(z)dz = 0$$

where $\gamma=L(-R,-r)\oplus - S(0,r)\oplus L(r,R) \oplus S(0,R)$

I have found that f(z) has a singularity at $0$ with a pole of order 1 and that Res(f,0)= -2i. where can I go from here?

asked Apr 29 '21 at 00:58

1

Let $I=\int_{-\infty}^{\infty}\frac{sin^2(t)}{t^2}dt$. Note that $\sin^2x=\frac{1-\cos2x}{2}=\frac{1}{2}\Re(1-e^{2iz})$. Integral along propose closed contour $=0$ (there are no poles inside contour), so $\oint=I+I_r+I_R=0$. $I_R\to0$ at $R\to\infty$, so $I=-I_r$ at $r\to0$ (integral along small semicircle clockwise around $x=0$) – Svyatoslav Apr 29 '21 at 01:15
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https://math.stackexchange.com/questions/2092158/solve-int-0-infty-big-frac-sin-tt-big2-dt/2092222#2092222 – Apr 29 '21 at 01:53

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