# [SOLVED]Poles on the real line

#### dwsmith

##### Well-known member
$\int_{-a}^{a}\frac{dx}{\sqrt{a^2 - x^2}\big(x^2 + b^2\big)}, \quad a,b\text{ real}$
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

##### MHB Oldtimer
Staff member
$\int_{-a}^{a}\frac{dx}{\sqrt{a^2 - x^2}\big(x^2 + b^2\big)}, \quad a,b\text{ real}$
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 $$\displaystyle \int_{-\pi/2}^{\pi/2}\frac{d\theta}{a^2\sin^2\theta + b^2}$$. That can be integrated by elementary means and the result is $$\displaystyle \frac{\pi}{b\sqrt{a^2+b^2}}.$$

#### Random Variable

##### Well-known member
MHB Math Helper
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.