Is this a dirac delta function? I had this on an exam yesterday, and I'm not entirely convinced that the statement is true.
We were asked to show that the function $\delta (x) = \int_{-∞}^{∞} \frac{1}{t(t-x)} dt$ is a dirac delta function by demonstrating that $I=\int_{-∞}^{∞} f(x)\delta(x) dx$ holds all the necessary properties.
There are three things I believe should be shown: 1) The function should be infinite at a single point. (this function is infinite at $t=x$)  2)  It should be zero everywhere else 3) It should satisfy $\int_{-∞}^{∞} f(x)\delta(x) dx=f(0)$.  
I showed 2 is true by demonstrating that the Cauchy Principal Value is zero for that integral which means that it's zero everywhere save that one point we avoid.... but I don't see how 3 holds in general.  I see that it holds for some functions, like $f(x)=x$, but what about $f(x)=1$, for example.  Anyway, is this a delta function or not.... if so, why?

The question as posted is not well defined (due to multiple reasons). The fact that Dirac's $\delta$ is not a function but rather a distribution was pointed out before. What is also (maybe even more) relevant is  that there is no prescription given how to handle the divergency due to the poles at $t=0$ and $t=x$. In fact, the question can be read as: is $I$ defined via
$$I = \int_{-∞}^{∞} \frac{1}{(t+ i \epsilon_1) (t-x+ i \epsilon_2)} dt$$
where $\epsilon_1,\epsilon_2 \to 0$ the $\delta$-distribution. It turns out that the answer to this question depends on the fact whether $\epsilon_1$ (and $\epsilon_2$) approach 0 from above or from below. That this is important is exemplified by Sokhotski's formula.
To get more insight into the problem, we can apply the partial fraction expansion
$$\frac{1}{(t+ i \epsilon_1) (t-x+ i \epsilon_2)}
= \frac{1}{x+ i(\epsilon_1-\epsilon_2) } \left(\frac{1}{t-x+i\epsilon_2}- \frac{1}{t+ i \epsilon_1}
 \right).$$
If we perform the integral over $t$ (which is now possible as we have regularized the integral by the small imaginary parts), we obtain 
$$\int_{-\infty}^\infty \left(\frac{1}{t-x+i\epsilon_2}- \frac{1}{t+ i \epsilon_1}
 \right)dt = \begin{cases} 2\pi i & \epsilon_1 >0 , \epsilon_2 <0,\\
-2\pi i & \epsilon_1 <0 , \epsilon_2 >0,\\
0 & \text{else}.
\end{cases}$$
We see that if $\epsilon_1$ has the same sign as $\epsilon_2$, $I$ cannot be the $\delta$-distribution as it vanishes identically.
For $\epsilon_1 \epsilon_2 <0$, we have
$$I = \frac{2\pi i \mathop{\rm sgn}(\epsilon_1) }{x+ i(\epsilon_1-\epsilon_2) }
= 2\pi i \mathop{\rm sgn}(\epsilon_1) \left( \mathcal{P} \frac{1}{x} 
+ i \pi \mathop{\rm sgn}(\epsilon_2-\epsilon_1) \delta(x)
\right)$$
via Sokhotki's formula.
We see that the distribution defined by $I$ depends very sensitively on the signs and magnitude of $\epsilon_1$ and $\epsilon_2$. And in non of the many different limits it is in fact a simple $\delta$-distribution!
