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?


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

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


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

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

  • $\begingroup$ 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 $\endgroup$ – User001 Dec 1 '15 at 4:27
  • 1
    $\begingroup$ You're welcome!! As always, my pleasure. $\endgroup$ – Mark Viola Dec 1 '15 at 4:28
  • $\begingroup$ 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, $\endgroup$ – User001 Dec 1 '15 at 4:34
  • 2
    $\begingroup$ $y$ 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 $1/2$. $\endgroup$ – Mark Viola Dec 1 '15 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.