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How can i find $$\int_{-\infty}^{\infty} \frac{\sin x}{x}e^{ix}dx$$?

my attempt:

$$\int_{-\infty}^{\infty} \frac{\sin x}{x}e^{ix}dx$$$$=\int_{-\infty}^{\infty} \frac{\sin x}{x}(\cos x+i\sin x)dx$$ $$=\int_{-\infty}^{\infty} (\frac{\sin 2x}{2x}+i\frac{1-\cos2x}{2x})dx$$

i got stuck here. how can i proceed from here? please help me

Jyerki menon
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  • Maybe write sin(x) as $\frac{e^{ix}-e^{-ix}}{2i}$ – Mars Aug 30 '20 at 19:46
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    Why do you think integral converges? – openspace Aug 30 '20 at 19:48
  • I think this needs to be interpreted as an inverse Fourier transform. – Christoph Aug 30 '20 at 19:52
  • Taken as the principal value, we have $\int {-\infty}^{\infty} \sin^2(x)/x dx=0$ and $\int{-\infty}^{\infty} \sin(x)/x,dx= \pi$ is well-known; see, for instance, https://math.stackexchange.com/q/5248 – Integrand Aug 30 '20 at 19:55
  • @openspace. It isn't Lebesgue integrable over all of $\mathbb R,$ but the limit $\lim_{R\to\infty} \int_{-R}^{R} \frac{\sin x}{x} e^{ix} , dx$ should converge since the integrals $\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x} e^{ix} , dx$ form an alternating sequence of decreasing (in size) values. So the integral from $0$ to $\infty$ should converge as an alternating series. – md2perpe Aug 30 '20 at 19:56
  • @md2perpe I thought about convergence in P.V. But I even don't know what does the author want – openspace Aug 30 '20 at 19:57
  • https://math.stackexchange.com/a/3808628/168433 – md2perpe Aug 30 '20 at 20:22

1 Answers1

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Indeed, your solution is almost complete. Now with Dirichlet's integral, substitute $t=2x$:

$$\int_{-\infty}^{\infty} \frac{\sin 2x}{2x}dx=\frac{1}{2}\cdot\int_{-\infty}^{\infty} \frac{\sin 2x}{x}dx=\frac{1}{2}\cdot\int_{-\infty}^{\infty} \frac{\sin t}{t}dt=\frac{\pi}{2}$$ And the integral in the imaginary parts, substitute $t=-x$: $$\int_{-\infty}^{\infty}\frac{\sin^2 x}{x}dx=\int_{-\infty}^{0}\frac{\sin^2 x}{x}dx+\int_{0}^{\infty}\frac{\sin^2 x}{x}dx=-\int_{0}^{\infty}\frac{\sin^2 t}{t}dt+\int_{0}^{\infty}\frac{\sin^2 x}{x}dx=0$$ Hence: $$\int_{-\infty}^{\infty} \frac{\sin x}{x}e^{ix}dx=\int_{-\infty}^{\infty} \left(\frac{\sin 2x}{2x}+i\frac{\sin^2 x}{x}\right)dx=\int_{-\infty}^{\infty} \frac{\sin 2x}{2x}dx+i\int_{-\infty}^{\infty}\frac{\sin^2 x}{x}dx=\frac{\pi}{2}.$$

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