Compute $$\int_{-\infty}^{\infty} \frac{\cos(αx)}{(x^2+1)(x^2+4)} \mathrm dx. $$

DonAntonio gave a well-constructed answer here but I have a few confusions on his answer as I am new to complex integral:

  1. How did we derive $\text{Res}_{z=i}(f)=\lim_{z\to i}(z-i)f(z)=\frac{e^{-\alpha}}{2i\cdot 3i\cdot (-i)}=-\frac{e^{-\alpha}}6i\;\;$? (the step when solving the limit)
  2. What is the purpose of using Jordan's Lemma to get $\lim_{R\to\infty}\int\limits_{\gamma_R}f(z)dz=0$? The result equalling to 0 does not seem to be applied to the next equation.
  3. Should the final answer be $\frac16\left(2e^{-\alpha}-e^{-2\alpha}\right)\pi$ or without $\pi$?

Thanks in advance.


2 Answers 2

  1. Because\begin{align}\lim_{z\to i}(z-i)f(z)&=\lim_{z\to i}(z-i)\frac{e^{i\alpha z}}{(z+i)(z-i)(z^2+4)}\\&=\lim_{z\to i}\frac{e^{i\alpha z}}{(z+i)(z^2+4)}\\&=\frac{e^{-\alpha}}{(2i)\times3}\\&=-\frac{e^{-\alpha}}6i\end{align}
  2. You need the fact that $\lim_{R\to\infty}\int_{\gamma_R}f(z)\,\mathrm dz=0$ to deduce that$$\int_{-\infty}^\infty f(x)\,\mathrm dx=\lim_{R\to\infty}\int_{C_R}f(z)\,\mathrm dz.$$
  3. It's $2\pi i$ times the sum of the residues at $i$ and at $2i$. So, yes, $\pi$ should be there.
  • $\begingroup$ thank you so much, just one last question, for Felix Marin's similar answer in the same link, i do not understand how he derived his first line's first equal relationship? i.e. how is 1/(x^2+1)(x^+4) equal to 1/3(x^2+1)-1/6(x^-1)? $\endgroup$ Jan 25, 2021 at 0:04
  • $\begingroup$ Actually, what he claims is that$$\int_{-\infty}^\infty\frac{\cos(\alpha x)}{(x^2+1)(x^2+4)}\,\mathrm dx=\frac13\int_{-\infty}^\infty\frac{\cos(\alpha x)}{x^2+1}\,\mathrm dx-\frac16\int_{-\infty}^\infty\frac{\cos(2\alpha x)}{x^2+1}\,\mathrm dx.$$But, no, I don't know where is that this comes from. But Felix Marin is an active member, and so you can post a comment on his answer to ask for a clarification. $\endgroup$ Jan 25, 2021 at 7:27
  • $\begingroup$ understood thank you sir $\endgroup$ Jan 25, 2021 at 20:40

$$J=\int_{-\infty}^\infty\frac{\cos(\alpha x)}{(x^2+1)(x^2+4)}dx\tag{1}$$

$$\frac{1}{(x^2+1)(x^2+4)}=\frac13\left(\frac1{x^2+1}-\frac1{x^2+4}\right)$$ lets make a general integral: $$I(\alpha,\beta)=\int_{-\infty}^\infty\frac{\cos(\alpha x)}{x^2+\beta^2}dx$$ this defines your integral as: $$J=\frac{I(\alpha,1)-I(\alpha,2)}{3}\tag{2}$$

Lets use the Feynman technique: $$I_\alpha=-\int_{-\infty}^\infty\frac{x\sin(\alpha x)}{x^2+\beta^2}dx=-\int_{-\infty}^\infty\frac{(x^2+\beta^2-\beta^2)\sin(\alpha x)}{x(x^2+\beta^2)}dx$$ $$I_\alpha=-\int_{-\infty}^\infty\frac{\sin(\alpha x)}{x}dx+\beta^2\int_{-\infty}^\infty\frac{\sin(\alpha x)}{x(x^2+\beta^2)}dx$$ now notice that this first integral is just the dirichlet integral so we can say: $$I_\alpha=-\pi+\beta^2\int_{-\infty}^\infty\frac{\sin(\alpha x)}{x(x^2+\beta^2)}dx$$ now what we are going to do is differentiate wrt $\alpha$ again: $$I_{\alpha\alpha}=\beta^2\int_{-\infty}^\infty\frac{\cos(\alpha x)}{x^2+\beta^2}dx$$ but notice that this is just the same as saying: $$\frac{\partial^2}{\partial\alpha^2}I(\alpha,\beta)=\beta^2I(\alpha,\beta)$$ now this is just a pretty common pde which we can set up as the following system: $$I=I(\alpha,\beta)$$ $$I_{\alpha\alpha}-\beta^2I=0$$ $$I(0,\beta)=\int_{-\infty}^\infty\frac{dx}{x^2+\beta^2}=\frac{\pi}{\beta}$$ $$I_\alpha(0,\beta)=-\pi$$

I get: $$I(\alpha,\beta)=\frac{\pi}{\beta}e^{-\alpha\beta}$$ and so: $$J=\frac13\left(\pi e^{-\alpha}-\frac{\pi}{2}e^{-2\alpha}\right)=\frac{\pi(2e^{-\alpha}-e^{-2\alpha)}}{6}$$

So yes the $\pi$ should be there. Just a quick note to add: The equation works for $\beta\in\mathbb{R}^+/\{0\},\alpha\in\mathbb{R}^+$


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