Prove or disprove following integral. Assume $L$ a constant, and assume $x$ real. Is the following equation true?
$$
\int_{-\infty}^\infty\frac{1}{k^2}\exp(-ikx)dk = \frac{L}{|x|}
$$
If it is true, find the value of $L$. If it is not, show why is not.
 A: $$
I(x)=\int_{-\infty}^\infty\frac{1}{k^2}\exp(-ikx)dk 
$$
We recognize this as the inverse FT of $1/k^2$
Let's look at the second derivatice with respect to $x$:
$$
I''(x)=-\int_{-\infty}^\infty\exp(-ikx)dk 
$$
Note that from a formal point of view this step has to be handled with care, because we are dealing with divergent expressions, so exchanging integration and differentiation is not justified in the usual way. We have to refer to the theory of tempered distributions to make all steps rigorous.
In this sense the above integral is given by
$$
I''(x)=-2\pi\delta(x)
$$
where $\delta(x)$ denotes Dirac's Delta distribution.
Integrating this (in distributional sense) with respect to x we obtain
$$
I'(x)=\pi(2\Theta(x)-1)
$$
Here $\Theta(x)$ is the Heaviside step function.
The constant of integration is fixed in a way that we have a sign change at $x=0$ . That this is necessary can be seen from the definition of $I'(x)$ looking at its imaginary part. 
Another careful integration gives
$$
I(x)=-\pi |x|
$$
The last constant of integration is given by the requirement that $I(\infty)=0$.
This is again a step where one has to take some care.
We can conclude that your conjecture is wrong. 
It would be true if we would consider a $3\text{D}$ version of your integral. The constant would be $1$ if i'm not mistaken...
Edit:
As a reference that our formula is indeed correct: Look at formula 3.10.3 and take $n=2$ 
