Efficient Techniques for Solving Complex Integrals | Help with Definite Integral

[jazhemar]

I'm having a tough time with this integral:

$$\int_{0}^\infty \frac{x^2 \, dx}{x^4+(a^2+\frac{1}{b^2})x^2+\frac{2a^2}{b^2}}$$
where $$a, b \in \Bbb R^+$$ I tried using the residue theorem, but the roots of the denominator I found are quite complicated, and I got stuck.

What contour should I use? Is there an alternative method? I would appreciate any advice.

 

[Ackbach]

You'll probably need to assume that the imaginary parts of the roots are positive. I would go with the contour of negative infinity to positive infinity, and then do a semicircle back. That is, do this:
\begin{align*}
f(z)&=\dfrac{z^2}{z^4+(a^2+1/b^2)z^2+2a^2/b^2} \\
I&=\int_0^{\infty}f(z) \, dz \\
2I&=\int_{-\infty}^{\infty}f(z) \, dz, \qquad \text{because} \; f(z) \; \text{is even}.
\end{align*}
Then compute
$$\int_{-R}^R f(z) \, dz$$
along the real axis, and compute
$$iR\int_{0}^{\pi}f\Big(Re^{i\theta}\Big) \, e^{i\theta} \, d\theta,$$
where $z=R e^{i\theta}$. My hunch is that the semicircle part will be zero, and you can probably show that by the $ML$ inequality.

You are going to need to horse through the roots of the denominator, because you'll have
$$\lim_{R\to\infty}\left[\int_{-R}^Rf(z) \, dz+\int_{0}^{\pi}iR \,f\Big(Re^{i\theta}\Big) \, e^{i\theta} \, d\theta\right]=\sum_j\text{Res}[f(z_j),z_j],$$
and to find the residues, you'll need the roots of the denominator. They're really not that bad:
$$z^2=\frac{-a^2b^2-1\pm\sqrt{a^4b^4-6a^2b^2+1}}{2b^2}.$$

I think they're all simple roots, so you can compute the residues by
$$\text{Res}[f(z_j),z_j]=\lim_{z\to z_j}(z-z_j)f(z).$$

Can you continue?

 

[jazhemar]

Thank you! But, how do I know for which of those four roots are the imaginary parts positive?

 

[Ackbach]

Thank you! But, how do I know for which of those four roots are the imaginary parts positive?

So let's examine the equation
$$z^2=\frac{-a^2b^2-1\pm\sqrt{a^4b^4-6a^2b^2+1}}{2b^2},$$
which describes all four poles of $f(z)$. First of all, note that the discriminant
\begin{align*}
a^4b^4-6a^2b^2+1&=a^4b^4-2a^2b^2+1-4a^2b^2 \\
&=(a^2b^2-1)^2-4a^2b^2 \\
&=(a^2b^2-1-2ab)(a^2b^2-1+2ab) \\
&=(a^2b^2-2ab+1-2)(a^2b^2+2ab+1-2) \\
&=((ab-1)^2-2)((ab+1)^2-2).
\end{align*}
This might help you get a handle on where the roots lie, given the locations of $a$ and $b$.

Now, I don't know what the RHS is fully, but since the quadratic equation in $z^2$ has all real coefficients, then both of these solutions, if they are complex, come in complex-conjugate pairs. As a matter of fact, if any of these roots are real and positive, then the original integral blows up along its contour. I don't think this happens, though, and here's why:

\begin{align*}
(a^2b^2+1)^2&=a^4b^4+2a^2b^2+1 \\
&>a^4b^4-6a^2b^2+1,
\end{align*}
and hence
$$a^2b^2+1>\sqrt{a^4b^4-6a^2b^2+1}.$$
So, if the discriminant is positive, the square root of its magnitude is less than the negative thing being added to it. So when you take the positive sign, you get a negative, and when you take the negative sign, you get something even more negative. So there cannot be any positive real values of $z^2$.

I suppose there could be negative real roots, in which case the contours I suggested in my previous post would run right over them; you can either change your contour to be only in the first quadrant (You can't go on the imaginary axis, because a negative value of $z^2$ would square root to be on the imaginary axis! You'd need to do, say, from $0\le\theta\le\pi/4$), or you need to impose a condition on $a$ and $b$ such that this discriminant is negative, and you get only complex conjugate values for $z^2$.

So, let's suppose that is the case (and with the factors above, hopefully it will be a bit easier to find out when that happens): the roots of the quadratic in $z^2$ are complex conjugate pairs. That means $z^2=r e^{\pm i\theta}$ for some positive real $r$, and $0<\theta<\pi$. Finally, we have the four poles located at
$$z=\sqrt{r} \, e^{i\theta/2}, \; \sqrt{r} \, e^{-i\theta/2}, \; -\sqrt{r} \, e^{i\theta/2}, \; -\sqrt{r} \, e^{-i\theta/2}.$$
The first and last of these, if you think it through, will have positive imaginary part.

So, what is $\theta$? Well, we know that
$$\tan(\theta)=\frac{\text{Im}(z^2)}{\text{Re}(z^2)}.$$
Work out what that is, and you should be on your way. Don't forget to make sure $0<\theta<\pi$ by adding or subtracting $\pi$.