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Let $$f(z)=\int _{-\infty }^{\infty }\frac{\sin\left(wz\right)}{w}\:dw$$

$$f'(z)=\int _{-\infty }^{\infty }\sin\left(wz\right)\:dw=\lim _{x\to \infty }\left(\frac{\cos\left(wz\right)}{z}\right)-\lim _{x\to -\infty }\left(\frac{\cos\left(wz\right)}{z}\right)= \cos\left(\infty \right)-\cos\left(-\infty \right)=\cos\left(\infty \right)-\cos\left(\infty \right)=0$$

$f(z)= 0 + C$; $f(0)=\int _{-\infty }^{\infty }\frac{\sin\left(w0\right)}{w}\:dw \Leftrightarrow 0 = 0 + C \Leftrightarrow C=0 $

I should have obtained $\pi $, since $\frac{\sin\left(w\right)}{w}$ is an even function and $\int _{0\:}^{\infty \:}\frac{\sin\left(w\right)}{w}\:dw$ = $\frac{\pi }{2}$

Mårten W
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David Sousa
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  • what is $z$? in your function? – Guy Fsone Oct 11 '17 at 18:49
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    $\int_{-\infty}^\infty \sin(wz)dw$ diverges for every $z\ne 0$ you can't conclude this way. Try instead a change of variable. – reuns Oct 11 '17 at 18:50
  • he do not any right to differentiate through the integral – Guy Fsone Oct 11 '17 at 18:51
  • @reuns change of variable requires multivariable calculus right ? – David Sousa Oct 11 '17 at 18:52
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    No, change of variable just says if $F'(x) = f(x)$ and $G(y) = F(\phi(y))$ then $G'(y) = f(\phi(y))\phi'(y)$ so that $$\int_a^b f(\phi(y)) \phi'(y)dy = G(b)-G(a)= F(\phi(b))-F(\phi(a))= \int_{\phi(a)}^{\phi(b)} f(x)dx$$ – reuns Oct 11 '17 at 18:55

2 Answers2

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If $z\in \Bbb R$ then letting $u=wz$ with $z\neq 0$ implies $dw =\frac{du}{z}$

then

$$f(z)=\int _{-\infty }^{\infty }\frac{\sin\left(wz\right)}{zw}\:d(zw) =sign(z)\int _{-\infty }^{\infty }\frac{\sin\left(u\right)}{u}\:du \\=2sign(z)\int _{0}^{\infty }\frac{\sin\left(u\right)}{u}\:du=\begin{cases} \pi, & z > 0 \\ 0, & z = 0 \\ -\pi, & z < 0 \end{cases}$$

Where we used this Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?

Guy Fsone
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By change of variables :

If $z$ or $w$ $\in \mathbb R$, let $u = wz,z\neq 0$ from which you get : $dw =\frac{du}{z}$

$$\int _{-\infty }^{\infty }\frac{\sin\left(wz\right)}{zw}\:d(zw) =\int _{-\infty }^{\infty }\frac{\sin\left(u\right)}{u}\:du =2\int _{0}^{\infty }\frac{\sin\left(u\right)}{u}\:du=\pi$$

Hint for a different approach if you're over complex analysis :

If $z$ or $w \in \mathbb C$ , the way to solve this is complex integration.

$$\int _{-\infty }^{\infty }\frac{sin\left(wz\right)}{w}\:dw =\Re\bigg\{\int _{-\infty }^{\infty }\frac{e^{iwz}}{w}\:dw \bigg\}$$

Rebellos
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