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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?

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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 $\log{s^2}$

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