# Integrating with residues vs integral convergence

Let me cite a theorem and then ask a question

Let $f$ be meromorphic in a neighborhood of $\overline{H_{+}}$ with a finite number of poles which belong to $H_{+}$. If there exist $A>0,R>0,\epsilon>0$ such that $$|f(z)|\leq A|z|^{-\epsilon}$$ as long as $|z|>R$, then for $a>0$ $$\int_{-\infty}^{\infty}f(x)e^{iax}dx=2\pi i\sum_{a_{j}\in H_{+}}\text{res}(fe^{iaz};a_{j})$$

Now, take $\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+4}dx$ as an example. Taking $f(z)=\frac{z}{z^2+4}$, we see that $$|f(z)|=\frac{|z|}{|z^2+4|}\leq2\frac{1}{|z|}$$ for $|z|>8$. Our function satisfies the assumptions in the theorem and has only one pole in $H_{+}$. Taking the imaginary part of $2\pi i\cdot\text{res}(fe^{iz};2i)$ leads to the answer. Now, my question. Say we want to compute the integral $\int_{-\infty}^{\infty}\frac{P(x)}{Q(x)}\sin xdx$ such that $\text{deg}P=\text{deg}Q-1$ and $Q\neq 0$. If we do nothing but apply the theorem above, and by doing so we obtain some number, does it prove that the integral is convergent? In other words, if someone asked me about the convergence of that real integral and I showed him a solution involving the above theorem, would it be valid? I'm asking this question because in the past I've seen problems like: prove (using methods of calculus) that such an integral is convergent and then, using methods of residues, find its value.

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I think such integrals are convergent by Jordan's Lemma. en.wikipedia.org/wiki/Jordan's_lemma – Ron Gordon Feb 5 '13 at 8:43
@rlgordonma I don't think that's what the OP is asking. That the integral along the semicircle goes to zero does follow from Jordan's lemma. The OP is concerned about the convergence of the real integral. A semi-circular contour will just give you Cauchy's principal value of the integral. – mrf Feb 5 '13 at 11:36

By doing what you describe, you will show that

$$\lim_{R\to\infty} \int_{-R}^R f(x)e^{iax}\,dx$$

converges. This is usually called Cauchy's principal value of the integral, and may exist even if the improper integral doesn't exist.

The problem is that $\int_{-\infty}^{\infty} f(x)e^{iax}$ in general is not absolutely convergent unless you assume something like $|f(x)| < C|x|^{-1-\epsilon}$ and it's not obvious that the improper integral exists.

However If $f = p/q$ where $\deg q = 1+\deg p$, you can integrate over a rectangle with vertices at $a$, $a+iR$, $-b+iR$ and $-b$ instead of over a semi-circle. (In fact, the analog to Jordan's lemma is easier to prove for such contours!) By letting $a \to \infty$, $b\to \infty$ and $R\to\infty$ independently of each other, it follows that the improper integral

$$\int_{-\infty}^{\infty} f(x)e^{iax}\,dx = \lim_{\substack{a\to\infty \\ b\to\infty}} \int_{-b}^a f(x)e^{iax}\,dx$$

exists (and is equal to the correponding sum of residues)!

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