how to calculate integral of following function over infinity ? i tried to solve this in ever program and ask some of my teacher they can't answer please help me. $$\int_{-\infty}^\infty\frac{e^{ix}}{x^2}$$
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That integral does not exist. There are various ways of "interpreting" some divergent integrals - Cauchy principal value, summation methods, whatever. None of them work here; the singularity at the origin is "like" $1/t^2$, no cancellation available. The integral really really does not exist.
David C. Ullrich
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This a stretch, but perhaps you started with an expression that was meant to be interpreted only in a weak sense such as $$G(x,x')=\int_{-\infty}^{\infty} \frac{\mathrm{e}^{\mathrm{i}k(x-x')}}{k^2}\frac{\mathrm{d}k}{2\pi}$$ in which case we might say that $$G(x,x')=-\tfrac{1}{2}\lvert x-x'\rvert\text{,}$$ i.e., $$\int\frac{\mathrm{e}^{\mathrm{i}k}}{k^2}\mathrm{d}k\text{"$=$"}-\pi\text{.}$$
K B Dave
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I know what a weak solution to a differential equation is. I don't see how citing a link to that converts a clearly divergent integral to $-|x-x'|$. – David C. Ullrich Apr 09 '21 at 19:38
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We can solve this with a regulator-based approach (see also here). Let
$$ f(\eta) = \mathrm{e}^{-\eta} (1 - 2 \eta) $$
be our regulator. Then
\begin{align} \int_{-\infty}^\infty \frac{\mathrm{e}^{ix}}{x^2} \mathrm{d}x &= \int_{-\infty}^\infty \lim_{\varepsilon \rightarrow 0} \frac{\mathrm{e}^{ix}}{x^2} f\left(\left(\frac{2 \varepsilon}{x}\right)^2\right) \mathrm{d}x \\ &\stackrel{!}{=} \lim_{\varepsilon \rightarrow 0} \int_{-\infty}^\infty \frac{\mathrm{e}^{ix}}{x^2} f\left(\left(\frac{2 \varepsilon}{x}\right)^2\right) \mathrm{d}x \\ &= \lim_{\varepsilon \rightarrow 0} \left( -\frac{4 \pi \varepsilon^2}{3} {_0}F_2\left(;2,\frac{5}{2};\varepsilon^2\right) - \pi {_0}F_2\left(;1,\frac{3}{2};\varepsilon^2\right) + \frac{\sqrt{\pi}}{2 \varepsilon} \left({_0}F_2\left(;\frac{1}{2},\frac{1}{2};\varepsilon^2\right) - {_0}F_2\left(;-\frac{1}{2},\frac{1}{2};\varepsilon^2\right) \right)\right) \\ &= 0 - \pi + 0 \\ &= -\pi \end{align}
where where ${_p}F_q(a;b;z)$ is the generalized hypergeometric function.
user76284
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@DavidC.Ullrich No problem. To be clear, the original integral is of course divergent. I'm talking about its regularized value (hence the word 'regulator'). – user76284 Apr 09 '21 at 19:52
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@user76284: As David C Ulrich pointed out, the integral does not exists in the sense of Lebesgue, not even in the sense of distributions. $u=1/x^2$ is understood to act on $\mathcal{S}$ as $$u:\phi\mapsto \lim_{\varepsilon\rightarrow0}\Big(\int_{|x|>\varepsilon}\frac{\phi(y)}{y^2},dy -2\frac{\phi(0)}{\varepsilon}\Big)$$
In that sense, its Fourier transformis given by $\widehat{u}(\phi)=-\pi\int\operatorname{sign}(x)x\phi(x)$
– Mittens Apr 09 '21 at 22:14