How can one solve this integral analytically? Integrals of the following form show up frequently in Physics, especially in electrodynamics. I have the following integral that I have no idea how to approach solving it:
$$\int_{-\infty}^{\infty}\left[\frac{1}{\sqrt{(x-x')^2+(y-a)^2+z^2}}-\frac{1}{\sqrt{x'^2+a^2}}\right]dx'.$$
I can solve each of them separately, or look them up in table integrals, but, the caveat is that separately, the integrals diverge, but as a whole, the integral converges. The only way I have been able to solve this is via Walfram Alpha and only with numbers--that is, replacing $z^2$ by 4 say and trying to figure out what goes where in the final answer. 
I would love for someone to guide me as to how I can solve this integral exactly on paper. Additionally, could someone explain why the integral diverges when analyzed separately compared to when analyzed as a whole? This seems to break the general rule:
$$\int_a^b[f(x)+g(x)]dx=\int_a^bf(x)dx+\int_a^bg(x)dx.$$
[Perhaps some condition such as smoothness or continuity of functions is being broken in the presented case, but I am not sure.]
 A: To find $$\int \frac{1}{\sqrt{u^2+a^2}} du$$ (a > 0), let $x = a \tan \theta$,  Then $dx = a \sec^2 \theta d \theta$, and $\sqrt{u^2+a^2} = a \sec \theta$ (here we are using the identity $\sec^2 \theta = 1 + \tan^2 \theta$).  This transforms the integral to $$\int \frac{a \sec^2 \theta}{a \sec \theta} d \theta = \int \sec \theta d \theta = \ln | \sec \theta + \tan \theta| = \ln | \frac{\sqrt{u^2+a^2}}{a}+\frac{u}{a}| + C.$$ Furthermore, properties of logs allow as to slurp the $a$'s in the denominator into the $+C$, so $$\int \frac{1}{\sqrt{u^2+a^2}} du = \ln | \sqrt{u^2+a^2}+u|+C$$
Now if we let $b = \sqrt{(y-a)^2+z^2}$, this and a simple substitution gives that your indefinite integral is equal to $$\ln | \sqrt{(x'-x)^2+b^2}+x'-x|-\ln|\sqrt{x'^2+a^2}+x'| + C = \ln|\frac{\sqrt{(x'-x)^2+b^2}+x'-x}{\sqrt{x'^2+a^2}+x'}| +C.$$  So if $f(x')$ is this expression, we want $\lim_{x' \to \infty} f(x') - \lim_{x' \to -\infty} f(x')$.  It's a standard Calc I exercise to show that $\lim_{x' \to \infty} f(x') = \ln(1)=0$.  To find $\lim_{x' \to -\infty} f(x')$ multiply the top and bottom by the conjugates of both the top and bottom to get $$f(x') = \ln|\frac{b^2}{a^2} \frac{\sqrt{x'^2+a^2} - x'}{\sqrt{(x'-x)^2+b^2}-(x'-x)}|.$$  Again the Calc I technique shows that $\lim_{x' \to -\infty} f(x') =  \ln\frac{b^2}{a^2}$.
So your over answer is $-\ln(\frac{b^2}{a^2}) = 2 \ln(\frac{a}{b})= 2 \ln \frac{a}{\sqrt{(y-a)^2+z^2}}$.
To answer your second question, improper integrals are defined in terms of limits, and the limit law $$\lim_{x \to a} (f(x)+g(x))=\lim_{x\to a} f(x)  + \lim_{x \to a} g(x)$$ only holds when both $\lim_{x\to a} f(x)$ and $\lim_{x\to a} g(x)$ exist.
A: FOREOWRD: this answer doesn't represent a proper answer to the above question because instead of $x'^2 + a^2$ I wrote $x'^2 - a^2$. Yet I find it "cool and useful" and I will leave it here the same. I will however provide for another answer about the second terms of the integral.
Before a numerical analysis, it's better to rewrite it in another way.
Let's define:
$$(x-x') = p$$
$$(y-a)^2 + z^2 = k^2$$
$$x'^2 - a^2 = (x'+a)(x'-a) = (x-p+a)(x-p-a) = -(p-a+x)(-p-a-x) = p^2 - c^2$$
Where $c = a+x$
Hence
$$-\int_{-\infty}^{+\infty} \left[\frac{1}{\sqrt{p^2 + k^2}} - \frac{1}{\sqrt{p^2-c^2}}\right]dp$$
The indefinite integrals, separately, are trivial:
$$\int \frac{1}{\sqrt{p^2 + k^2}}\ dp = \log \left(\sqrt{k^2+p^2}+k\right)$$
$$\int \frac{1}{\sqrt{p^2-c^2}}\ dp = \log \left(\sqrt{k^2+p^2}+p\right)$$
Which shows you why they are separately divergent.
The the indefinite integral of them both is:
$$-\int \left[\frac{1}{\sqrt{p^2 + k^2}} - \frac{1}{\sqrt{p^2-c^2}}\right]dp = \log \left(\sqrt{p^2-c^2}+p\right) - \log \left(\sqrt{k^2-c^2}+p\right)$$
The problem now is: What are $c$ and $k$? 
From how we defined it, $k$ is surely always positive, and it also doesn't cause problems.
But $c$, well $c = a+x$, and we don't know if it's always positive. Also it appears as $-c^2$ hence $p^2 - c^2 \geq 0$ is required to use log unifying property.
We also need $p$ such that the two logarithm arguments will be strictly positive.
Why am I telling you all this? Because log of negative argument "exist" in the comple field, but we couldn't use the usual properties. This is why we cannot say anything a propri, unless we make strong assumptions.
Anyway, assuming it is all good, we can write the result as
$$\log\left(\frac{\sqrt{p^2 - c^2} + p}{\sqrt{p^2 + k^2} + p}\right)$$
Evaluating it as $p\to \pm \infty$ we easily find
$$\to 0 ~~~~~~~ \text{for} ~~ p\to +\infty$$
$$\to \log \left(-\frac{c^2}{k^2}\right) ~~~~~~~ \text{for} ~~ p\to -\infty$$
The final result is anyway complex.
