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Is it possible to get some form of $\delta$ function?

Just like $\int_{-\infty }^{\infty } e^{ 2\pi ik x} dk$ gives $\delta(x)$.

Sorry for not being clear before editing.

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  • $\begingroup$ Hve you tried using the Residue Theorem? $\endgroup$ – MathematicianByMistake Mar 21 '16 at 11:30
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    $\begingroup$ what sense do you give to your divergent integral ? as the limit of what ? and $\displaystyle\frac{1}{i \pi \xi}$ is the Fourier transform of $\text{sign}(x)$ $\endgroup$ – reuns Mar 21 '16 at 11:32
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    $\begingroup$ The integral does not exist. Consider the real part around 0. $\endgroup$ – PhoemueX Mar 21 '16 at 11:34
  • $\begingroup$ Looks similar to the Exponential integral? $\endgroup$ – Simply Beautiful Art Mar 21 '16 at 11:40
  • $\begingroup$ It could still have a Cauchy principal value. $\endgroup$ – Henning Makholm Mar 21 '16 at 11:41
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The integral does not exist, but it's Cauchy principal value does. To compute the PV, consider the following integral in the complex plane:

$$\oint_C dz \frac{e^{i z}}{z} $$

where $C$ is a semicircle of radius $R$ in the upper half plane with a small semicircular detour of radius $\epsilon$ about the origin. The integral is equal to

$$\int_{-R}^{-\epsilon} dk \, \frac{e^{i k}}{k} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{e^{i \epsilon e^{i \phi}}}{\epsilon e^{i \phi}} \\ + \int_{\epsilon}^R dk \, \frac{e^{i k}}{k} + i R \int_{0}^{\pi} d\theta \, e^{i \theta} \frac{e^{i R e^{i \theta}}}{R e^{i \theta}}$$

As $R \to \infty$, the fourth integral vanishes. We know this because its magnitude is bounded by

$$2 \int_0^{\pi/2} d\theta \, e^{-R \sin{\theta}} \le 2 \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi}\le \frac{\pi}{R}$$

Here we used the fact that $\sin{\theta} \gt \frac{2 \theta}{\pi}$ when $ \theta \in [0,\pi/2]$.

As $\epsilon \to 0$, the second integral does not vanish but approaches $-i \pi$. The first and third integrals combine to form the Cauchy PV of the original integral.

By Cauchy's theorem, the contour integral in the complex plane is zero. Thus,

$$PV \int_{-\infty}^{\infty} dk \, \frac{e^{i k }}{k} = i \pi$$

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Evaluating the Cauchy principal value.

If we have $f(x)=\frac{e^{ix}}{x}$, I note that $\Im f(-x)=\Im f(x)$ and $\Re f(-x)=-\Re f(x)$

If we also have $F(x)=\int f(x)dx$, we can see that there is a flip: $\Im F(-x)=-\Im F(x)$ and $\Re F(-x)=\Re F(x)$

Since $\int_{-a}^af(x)dx=F(a)-F(-a)$, we see that $\Im F(a)-F(-a)=2F(a)$ and that $\Re F(a)-F(-a)=0$.

Now it is just a matter of viewing the imaginary part of the problem.

We can see that $\Im f(x)=\frac{\sin(x)}{x}$.

The power series of this is given as

$$\frac{\sin(x)}{x}=\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots}{x}$$

$$=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\dots$$

$$\int\Im f(x)dx=x-\frac{x^3}{3\cdot3!}+\frac{x^5}{5\cdot5!}-\frac{x^7}{7\cdot7!}\dots$$

$$=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)(2n+1)!}$$

Now we just need to evaluate the limit to infinity to find the solution for the imaginary part of the integral.

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