1
$\begingroup$

Use residue calculus to compute the integral $\int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz$

My solution

If we add to the interval $I_{R}=[-R,R]$ add the semicircle $\gamma_{R}$ in the upper half plane with centre at the origin and radius $R$, then we obtain a closed contour $\Gamma_{R}$ (which we consider to be oriented in the positive direction over which the integral $f(z)=\frac{1}{(z^{2}+25)(z^{2}+16)}$ can be computed using the Cauchy's Residue theorem.

\begin{align} \int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz&=\lim_{R \to \infty}\int_{\Gamma_{R}}f(z)dz \\ &=2\pi i \sum_{k=1}^{2}\text{Res}(z_{k}), \quad \Im(z_{k})>0 \end{align} In this case, we have two simple poles at the points $z_{1}=4i$ and $z_{2}=5i$. The residues are computed as \begin{align} \text{Res}(z_{1})&=\frac{1}{(z^{2}+25)D(z^{2}+16)}\big|_{z=4i} \\ &=\frac{1}{(z^{2}+25)2z}\big|_{z=4i} \\ &=\frac{1}{9\cdot8i} \\ &=\frac{1}{72i} \\ &=-\frac{i}{72} \end{align} Similarly, we get for $z_{2}$ that \begin{align} \text{Res}(z_{2})&=\frac{1}{D(z^{2}+25)(z^{2}+16)}\big|_{z=5i} \\ &=\frac{1}{2z(z^{2}+16)}\big|_{z=5i} \\ &=\frac{1}{10i \cdot (-9)} \\ &=-\frac{1}{90i} \\ &=\frac{i}{90} \end{align} Thus, we get \begin{align} \int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz &=2\pi i \left(\text{Res}(z_{1})+\text{Res}(z_{2})\right) \\ &=2 \pi i \left(-\frac{i}{360}\right) \\ &=\frac{\pi}{180} \end{align}

Are there other methods one could use to arrive at this answer?

$\endgroup$
2
  • 2
    $\begingroup$ It can be solved using real analysis, by simply writing it as partial fractions and using the arctangent. $\endgroup$ Oct 3, 2014 at 20:40
  • $\begingroup$ Notice that, in order to use contour integrals, you must prove that the integral over the semi-circle vanishes as $R\to\infty$... $\endgroup$
    – bartgol
    Oct 3, 2014 at 20:52

1 Answer 1

0
$\begingroup$

Yes. Simply rewrite $$\int \frac{1}{(z^2+16)(z^2+25)} \mathrm{d}z= -\frac{1}{225} \int \frac{1}{\frac{z^2}{25}+1} \mathrm{d}z +\frac{1}{144} \int \frac{1}{\frac{z^2}{16}} \mathrm{d}z$$ and substitute $u=\frac{z}{5}$ for the first integral and $v=\frac{z}{4}$ for the second.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .