<s>[INST] Integral of $\frac{1}{\sqrt{(z-z')^2 + s^2}}$ I have a question about the signs of the antiderivative when one integrate $\frac{1}{\sqrt{(z-z')^2 + s^2}}$.
According to Wolfram Alpha here and here:
If one evaluates $\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz$, they get $\log (z-z' + \sqrt{(z-z')^2 + s^2}) + C$.
Evalutating $\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz'$ gives $-\log (z-z' + \sqrt{(z-z')^2 + s^2}) + C$.
However, one could rearrange the integrand before evaluating the second expression. Using the fact that $(z-z')^2 = (z'-z)^2$, one gets
\begin{equation}
\int \frac{1}{\sqrt{(z-z')^2 + s^2}} dz'= \int \frac{1}{\sqrt{(z'-z)^2 + s^2}} dz'
\end{equation}
which has the same form as the first integral, implying that the two should be equal.
What am I missing here?
 [/INST] Notice that
\begin{align}
-\log{(z-z'+\sqrt{(z-z')^2+s^2})} &= \log{\left(\frac{1}{\sqrt{(z-z')^2+s^2} -(z'-z)}\right)} \\
&= \log{\left(\frac{\sqrt{(z-z')^2+s^2} +(z'-z)}{(\sqrt{(z-z')^2+s^2} -(z'-z))(\sqrt{(z-z')^2+s^2} +(z'-z)}\right)} \\
&= \log{\left(\frac{\sqrt{(z-z')^2+s^2} +(z'-z)}{s^2}\right)} \\
&= \log{(\sqrt{(z-z')^2+s^2} +z'-z)} -\log{s^2} \\
\end{align}
So the two are equal except for a constant factor $\log{s^2}$
</s>