# Would a keyhole contour be advisable to use for this integration?

The integral is

$$\int_0^{\infty}\frac {1}{\sqrt{x}(1+x^2)}dx$$

which is to be evaluated by contour integration.

So, the integrand clearly has simple poles at $+/- i$.

But what kind of pole does the factor $\large \frac{1}{\sqrt{z}}$ have? Should I... "round up" to 1, so that $z=0$ is also a simple pole?

If what I said about the pole at $z=0$ is ok, then would a keyhole contour be advisable to use? The smaller circle would go to zero - and touch the pole -so is this an issue?

Or is there a better / correct contour to use instead?

Thanks,

• Why not simply substitute $x=y^2$ and evaluate the integral $$\int_0^\infty \frac{2}{1+x^4}\,dx=\int_{-\infty}^\infty\frac{1}{1+x^4}\,dx$$ – Mark Viola Dec 1 '15 at 4:12
• @LaplacianFourier The singularity at $z=0$ is a branch point, not a pole. – Mark Viola Dec 1 '15 at 4:15
• Such an awesome comment, @Dr.MV. -- thanks so much :-) – User001 Dec 1 '15 at 4:26

We can enforce the substitution $x\to x^2$ and write the integral of interest, $I$, as

$$I=\int_{-\infty}^\infty \frac{1}{1+x^4}\,dx \tag 1$$

The integral can be evaluated in terms of its residues in the upper-half plane as

\begin{align} I&=2\pi i\left(\text{Res}\left(\frac{1}{1+z^4}, z=e^{i\pi/4}\right)+\text{Res}\left(\frac{1}{1+z^4}, z=e^{i3\pi/4}\right)\right)\\\\ &=2\pi i \left(\frac{1}{4e^{i3\pi/4}}+\frac{1}{4e^{i9\pi/4}}\right)\\\\ &=\frac{\pi\sqrt{2}}{2} \end{align}

NOTE:

If one wishes to proceed using a keyhole contour, then we have

\begin{align} 0&=2\int_0^\infty \frac{1}{\sqrt{x}(1+x^2)}\,dx+2\pi i\text{Res}\left(\frac{1}{\sqrt{z}(1+z^2)},z=i\right)\\\\ &+2\pi i \text{Res}\left(\frac{1}{\sqrt{z}(1+z^2)},z=-i\right)\\\\ \int_0^\infty \frac{1}{\sqrt{x}(1+x^2)}\,dx&=-\pi i\left(\frac{1}{\sqrt{e^{i\pi/2}}(2i)}+\frac{1}{\sqrt{e^{i3\pi/2}}(-2i)}\right)\\\\ &=\frac{\pi\sqrt{2}}{2} \end{align}

• I will not look at your answer just yet :-). Will proceed now that you gave that awesome hint in the comments above. Thanks @Dr. MV – User001 Dec 1 '15 at 4:27
• You're welcome!! As always, my pleasure. – Mark Viola Dec 1 '15 at 4:28
• Hi @Dr.MV, in using the change of variable $x=y^2$, where x had ranged from 0 to infinity, how do we now know that $y^2$ ranges over the entire, extended real line? Shouldn't the integration variable, y, also range from 0 to $\infty$? Thanks, – User001 Dec 1 '15 at 4:34
• $y$ does range from $0$ to $\infty$. But then make use of the evenness of the integrand and extend the integral over the real line and multiply by $1/2$. – Mark Viola Dec 1 '15 at 4:37