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I want to prove that

$\delta(w) = \frac{1}{\pi^2} \int_{- \infty} ^{ \infty} \frac{dy}{y(y-w)}$

Could anyone help? I did the integration in two parts: $w=0$ and $w$ is not zero and I showed that for $w=0$, integral becomes infinite and for $w$ is not equal to zero it becomes zero. But I don't know why $\frac{1}{\pi^2}$ is present in the question. Could anyone add a better answer rather than mine?

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    In what sense do integrate? Principal value? Because even for $w\ne0$ this integral does not exist in the usual sense – 2017-02-05

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It comes from the Hilbert transform but you have to be very careful about how you define the integrals. You define:

$$ H(u)(t) = \frac{1}{\pi} {\rm P.V.} \int \frac{u(\tau)}{t-\tau} \; d\tau$$

A remarkable (and non-trivial) identity is that $H(H(u))(s)=-u(s)$ and your expression amounts to evaluating this for $s=0$. The identity $H\circ H=-{\rm Id}$ is valid in $L^p$, $1< p<+\infty$ but as you evaluate at a point, I suspect you need $u$ continuous or better.

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    what are $P$ and $L^p$ ? – 2017-02-06
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    If you are referring to P.V, it means "principal value" meaning that you integrate over $]-\infty,t-\epsilon[\cup ]t+\epsilon,+\infty[$ and let $\epsilon$ go to zero. And $L^p=L^p({\Bbb R})$. What is the context in which you have encountered this question? (lectures on distribution theory, functional analysis, or other ...?) – 2017-02-06
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    Mathematical methods for physicist, by Arfken – 2017-02-08
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    Ok, if you have learned about Fourier transform (and the fact that it takes convolution into products) then you may show (taking suitable limits) that $({\cal F} H u) (w) = i ({\cal F} u)(w) \sigma(w)$ where $\sigma(w)$ is the sign of $w$. Then ${\cal F} H \circ H (u)(w) = (i)^2 {\cal F} u = - {\cal F} u(w)$ implies by taking inverse Fourier transform $H H u = -u$ as I mentioned above. – 2017-02-08