MHB Poles on Real Line: Is \(z = \pm a\) Included?

[Dustinsfl]

\[
\int_{-a}^{a}\frac{dx}{\sqrt{a^2 - x^2}\big(x^2 + b^2\big)},
\quad a,b\text{ real}
\]
So we have poles at \(z = \pm ib\) but are the poles at the integration boundaries included at \(z = \pm a\)?

Should I have
\[
2\pi i \lim_{z\to b}(z - ib)\frac{1}{\sqrt{a^2 - z^2}(z^2 + b^2)} = \frac{\pi}{b\sqrt{a^2 + b^2}}
\]
or if I use the poles at the end points by taking pi i poles real + 2pi i UHP, I get the same answer as above. However, in matheatica, I get
\[
\frac{i\pi}{b\sqrt{a^2 - b^2}}
\]
so I am missing an \(i\) and I have + instead of a -.

 

[Opalg]

\[
\int_{-a}^{a}\frac{dx}{\sqrt{a^2 - x^2}\big(x^2 + b^2\big)},
\quad a,b\text{ real}
\]
So we have poles at \(z = \pm ib\) but are the poles at the integration boundaries included at \(z = \pm a\)?

Should I have
\[
2\pi i \lim_{z\to b}(z - ib)\frac{1}{\sqrt{a^2 - z^2}(z^2 + b^2)} = \frac{\pi}{b\sqrt{a^2 + b^2}}
\]
or if I use the poles at the end points by taking pi i poles real + 2pi i UHP, I get the same answer as above. However, in matheatica, I get
\[
\frac{i\pi}{b\sqrt{a^2 - b^2}}
\]
so I am missing an \(i\) and I have + instead of a -.

The integral of a real function over a real interval cannot possibly be imaginary. So the Mathematica answer is obviously rubbish.

I don't see where poles come into this at all. If you make the substitution $x=a\sin\theta$ then the integral becomes $$\int_{-\pi/2}^{\pi/2}\frac{d\theta}{a^2\sin^2\theta + b^2}$$. That can be integrated by elementary means and the result is $$\frac{\pi}{b\sqrt{a^2+b^2}}.$$

 

[polygamma]

The points $z = \pm a$ are not poles. They're branch points.

To evaluate the integral using contour integration the proper contour would be a dogbone/dumbell contour where the branch cut is on $[-a,a]$.

The fact that it equals $ \displaystyle 2 \pi i \ \text{Res} \Big[\frac{1}{\sqrt{a^{2}-z^{2}}(z^2+b^2)},ib\Big]$ is the result of the residue at infinity being zero.