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$$\int_0^\infty \frac{\sin w}w \, \cos xw \, dw$$

How can I solve this integral,I was thinking it may be solved by using Fourier transform, but It seems that it wouldn't work.besides I've tried many exchanging variable it wasn't helpful too.Do you have any idea?

mopy
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MAh2014
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    I think it would help to tackel this integral by using contour integration. Have you ever done this? – B. Pasternak Oct 29 '15 at 13:26
  • indeed Fouriertransforms are not the badest idea here. Your integral is just the real part of the FT of the $\text{sinc}$ function – tired Oct 29 '15 at 13:33
  • @B.Pasternak I haven't heard about that.could you elaborate please? – MAh2014 Oct 29 '15 at 13:34
  • Look up Fourier transform sinc – tired Oct 29 '15 at 13:35
  • @tired What about some integrals we don't know which function FT they are? – MAh2014 Oct 29 '15 at 13:37
  • http://math.stackexchange.com/questions/25903/derive-fourier-transform-of-sinc-function?rq=1 – tired Oct 29 '15 at 13:56
  • It seems that Fourier transforms are a better idea, but if you havent't heard about it, it might not be the best approach. The idea of a contour integral is that you extend the integrand to an analytic function defined on the complex plane, so that you can turn your integral into a complex line integral and use the Residue Theorem to magnificently simplify your integral, since you only have to deal with the poles inside your contour. But again, tired seems to give an approach mor suitable for you. – B. Pasternak Oct 29 '15 at 14:03

2 Answers2

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You can use the sine addition formula and get $$\sin w \cos xw = {1 \over 2}\sin(w + xw) + {1 \over 2}\sin(w - xw)$$ Hence your integral is equal to $${1 \over 2} \int_0^{\infty} {\sin (1 + x)w \over w} \, dw+ {1 \over 2} \int_0^{\infty} {\sin (1 - x)w \over w}\, dw $$ Substituting $u= (1 + x)w$ in the first and $u = (1 - x) w$ in the second gives (when $x \neq 1$ or $-1$) $${1 \over 2} sgn(1 + x)\int_0^{\infty} {\sin u \over u} \, du + {1 \over 2}sgn(1 - x) \int_0^{\infty} {\sin u \over u} \,du$$ $$= sgn(1 + x){\pi \over 4} + sgn(1 - x){\pi \over 4}$$ So when $|x| > 1$, the integral is zero, and when $|x| < 1$, the integral is ${\displaystyle {\pi \over 2}}$. The cases $x = \pm 1$ can be done directly: Here the integral is $${1 \over 2} \int_0^{\infty} {\sin 2w \over w} \, dw$$ Letting $u = 2w$ gives ${\displaystyle {\pi \over 4}}$ for these two cases.

Those familar with Fourier analysis might recognize these formulas, but the above shows them directly.

Zarrax
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I believe it is an equivalent statement if you take the derivative and integrate with respect to x. $$\int_0^\infty \frac{sin(w)}{w} cos(wx)dw=\int \frac{d}{d x}[\int_0^\infty \frac{sin(w)}{w} cos(wx)dw] dx$$ $$\rightarrow \int [\int_0^\infty \frac{sin(w)}{w} \frac{d}{d x}cos(wx)dw] dx$$ $$\rightarrow \int [\int_0^\infty \frac{sin(w)}{w} \frac{d}{d x}cos(wx)dw] dx$$ $$\rightarrow \int [\int_0^\infty \frac{sin(w)}{w} -wsin(wx)dw] dx$$ $$\rightarrow -\int [\int_0^\infty sin(w) sin(wx) dw] dx$$ Then expanding sin(a) as $\frac{e^{ia}-e^{-ia}}{2i}$: $$\rightarrow -\int [\int_0^\infty \frac{e^{iw}-e^{-iw}}{2i}\frac{e^{iw}-e^{-iw}}{2i} dw] dx$$ $$\rightarrow \frac{-1}{2i}\int [\int_0^\infty \frac{e^{iw(1+x)}-e^{-iw(1+x)}+e^{iw(1-x)}-e^{-iw(1-x)}}{2i} dw] dx$$ $$\rightarrow \frac{-1}{2i}\int [\int_0^\infty sin(w(1+x))+sin(w(1-x)) dw] dx$$ $$\rightarrow \frac{1}{2i}\int [ \frac{cos(w(1+x))}{(1+x)}+\frac{cos(w(1-x))}{1-x} |_0^\infty] dx$$ $$\rightarrow \frac{-1}{2i}\int \frac{1}{(1+x)}+\frac{1}{1-x} dx$$ $$\rightarrow \frac{-1}{2i}(ln(1+x)-ln(1-x)$$

Mark
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  • i think this is not true: The result should be $$ \frac{1}{4} \pi (\text{sgn}(1-x)+\text{sgn}(x+1))$$ – tired Oct 29 '15 at 13:50
  • What is a in your statement? And does differentiation under the integral not work because the singular points are included in the integration? – Mark Oct 29 '15 at 13:52
  • yeah this might cause problems, you have to think about distributions instead of functions – tired Oct 29 '15 at 13:53
  • @Mark You have some mistakes in multiplying.its result is$ \rightarrow \frac{-1}{2i}\int [\int_0^\infty \frac{e^{iw(1+x)}+e^{-iw(1+x)}-(e^{iw(1-x)}+e^{-iw(1-x)})}{2i} dw] dx$ – MAh2014 Oct 29 '15 at 13:57