A recent question from Juan Saloman reminded me of something that has nagged me for years, and I have never understood and never heard explained. (or maybe I just don't remember, but anyway …) In the complex plane, can an integral of the general form $\int dz f(z)$ over the complex plane be treated as a straightforward 1-D integral if no path in the complex plane is specified? I have been in classes where the professors seemed to do just that. But, I mean, $z$ has two independent parts, right? In other words, it's the complex plane, not the complex line. I have seen these kinds of integrals before and sort learned to deal with them in a monkey-see-monkey-do fashion, but never really understood what was going on to my own satisfaction. Thanks.
[Math] Is an integral in the complex plane an integral over a single number
complex integrationcomplex-analysis
Related Solutions
Briefly, holomorphic functions are immensely useful because, on the one hand, they are surprisingly common (since any power series, for example, whose coefficients grow reasonably slowly defines a holomorphic function), and on the other hand one can prove very strong theorems about them. There is a web of results including Cauchy's integral formula and the identity theorem which assert that holomorphic functions are astonishingly rigid: given information about a holomorphic function in a very small part of its domain, one can extract information about the function's behavior in other a priori unrelated parts of their domain (and this is what allows things like contour integration to work).
For that reason, holomorphic functions are a powerful tool to apply to a problem when they do apply. For example, analytic number theorists frequently construct holomorphic or meromorphic functions which carry number-theoretic information, such as the Riemann zeta function, to prove theorems like the prime number theorem. Since you say you have a background in combinatorics, you might enjoy reading Flajolet and Sedgewick's Analytic Combinatorics, a thorough exposition of (among other things) ways to use complex analysis to provide asymptotics for combinatorial sequences.
Here is a simple example. Let $f_n$ denote the number of ways that $n$ horses can win a race, with ties. It turns out that this sequence has generating function
$$F(z) = \sum_{n \ge 0} \frac{f_n}{n!} z^n = \frac{1}{2 - e^z}.$$
This function is meromorphic with poles at $z = \log 2 + 2 \pi i k, k \in \mathbb{Z}$, each of which has residue $-\frac{1}{2}$. In fact, it turns out that $F(z)$ admits an infinite partial fraction decomposition
$$F(z) = \sum_{k \in \mathbb{Z}} \frac{1}{2(\log 2 + 2 \pi i k - z)}.$$
And by expanding the terms on the RHS in a geometric series, this gives an asymptotic expansion for $\frac{f_n}{n!}$ with leading term $\frac{1}{2 (\log 2)^{n+1}}$. In other words,
$$f_n \sim \frac{n!}{2 (\log 2)^{n+1}}.$$
The pole at $z = \log 2$ dominates the asymptotic expansion: the leading term in the error of the above expression is given by the other poles nearest the origin, which occur at $z = \log 2 \pm 2 \pi i$. Because these poles have nonzero imaginary part, if you plot the error in the above approximation you'll find that it oscillates. It is not so easy to explain why this should be the case without complex analysis.
A famous example is Hardy and Ramanujan's asymptotic formula for the partition function
$$p(n) \sim \frac{1}{4n \sqrt{3}} e^{ \pi \sqrt{ \frac{2n}{3} } }$$
which is proven using a much more sophisticated version of the above argument.
But really, there is too much to say about holomorphic functions, so again I suggest that you read a textbook. Besides Needham's book, I also personally enjoyed Stein and Shakarchi, which is very user-friendly and has good applications.
You can actually see directly that there is no elementary expression for the integral, using contour integration. For your integral is the same as $$\mathrm{Im}\int_0^\infty {e^{iax} \over x^2 + b^2}\mathrm dx$$ Note that $${1 \over x^2 + b^2} = {1 \over 2ib}\left({1 \over x - ib} - {1 \over x + ib}\right)$$ So it suffices to evaluate the integrals $$\int_0^\infty {e^{iax} \over x \pm bi}\,\mathrm dx$$ These integrals converge despite the denominator having only degree one because the exponential factor modulates it, similar to for $\dfrac{\sin(x)}{x}$. You can use your quarter circle idea on each of these two terms (without even having to worry about indentations)... the upper quarter circle for $+bi$ and the bottom quarter circle for $-bi$. For example you get that the first one is equal to $$\int_0^\infty {e^{-ax} \over x + b}\,\mathrm dx$$ Letting $x = z - b$ this becomes $$e^{ab}\int_b^\infty {e^{-az} \over z}\,\mathrm dz$$ Then letting $z = \dfrac{w}{a}$ this turns into $$e^{ab}\int_{ab}^\infty {e^{-w} \over w}\,\mathrm dw$$ $$= e^{ab}\mathrm{Ei}(-ab)$$ Here $\mathrm{Ei}$ is the exponential function which is defined in terms of the above integral. So unless the exponential functions from the two quarter circles somehow cancel out (which I seriously doubt in view of Sasha's answer), you won't be able to get an elementary expression for the integral.
Best Answer
In the complex plane $\Bbb C$ certain functions $z\mapsto f(z)$ are distinguished as being holomorphic. Any such function comes with a domain $\Omega\subset\Bbb C$, which by definition is an open and connected set. Given such an $f$ and a curve $$\gamma:\ [a,b]\to\Omega, \quad t\mapsto z(t)\tag{1}$$ one can consider the integral $$\int_\gamma f(z)\ dz:=\int_a^b f\bigl(z(t)\bigr)\ z'(t)\ dt\ ,$$ whose value depends on $f$ as well as on $\gamma$.
When $f$ happens to be the derivative of some other analytic function $F:\ \Omega\to\Bbb C$, as $\ \cos\ $ is the derivative of $\ \sin$, and if $(1)$ is an arbitrary curve in $\Omega$ connecting the point $\gamma(a)=z_1\in\Omega$ with the point $\gamma(b)=z_2\in\Omega$ then $$\int_\gamma f(z)\ dz= F(z_2)-F(z_1)\ .\tag{2}$$ In order to prove $(2)$ consider the auxiliary function $$\phi(t):=F\bigl(z(t)\bigr)\qquad(a\leq t\leq b)$$ and rewrite the formula $\int_a^b \phi'(t)\ dt=\phi(b)-\phi(a)$ in terms of $f$ and $F$.
In this light the formula $\int_a^b f(x)\ dx=F(b)-F(a)$ from ordinary calculus can be considered as the special case where $\gamma$ is the directed segment $[a,b]\subset\Bbb C$.