# Integrals in the complex plane.

(Lecture 12 of Mathematical Methods II.)

The theory of integration received its first major mathematical formalisation through the Riemann integral (later enhanced in the form of Riemann-Stieltjes integral and again to attain its contemporary form with Lebesgue integrals), that is the limit of Riemann-sums:

$$S_k=\sum_{k=1}^N f(z_k)(z_k-z_{k-1})$$

but now with $z_k\in\mathbf{C}$. The meaning is not that of the surface enclosed by the fonction (as the result can be complex and the increment $dz=dx+i\,dy$ is itself also complex) but is nevertheless connected to an additive compound made from the function and weighted by small increments in the complex plane. Also, we have seen that in the case of harmonic functions, we could bring complex integrals in a form that express precisely the complex average of the function on a circle~(see here). We would find, for instance, $$\int_0^i z\,dz=\lim_{N\rightarrow\infty}\frac{i}{N}\Delta z+\frac{2i}{N}\Delta z+\cdots+\frac{Ni}{N}\Delta z$$ with $\Delta z$ one of the little path of integration as we go from $0$ to $i$ along the $y$ axis. The sum and limits are easily calculated as $\lim_{N\rightarrow\infty}\sum_{k=1}^N\frac{ik}{N}\frac{i}{N}=-1/2$.

Some extensions of this kind are straightforward, others are far reaching. We have seen already that much of the differential calculus takes the same form in the complex than in the real plane; for instance, rules of differentiation $(x^n)'=nx^{n-1}$ become simply $(z^n)'=nz^{n-1}$. Here too, formal results of this type, which are not particular of the real plane, will remain valid as well, so that, e.g., for $n$ positive, we will obtain:

$$\tag{1} \int z^n\,dz=\frac{z^{n+1}}{n+1}\,,\quad \int\frac{dz}{z^n}= \begin{cases} \displaystyle-\frac{1}{(n-1)z^{n-1}} & \text{if }n>1\,, \\ \log(z) & \text{if }n=1\,, \end{cases}$$

etc. In particular, $\int z\,dz=z^2/2$ and we would recover the above result by substituting boundaries. Also properties like $\int f+g=\int f+\int g$, etc., still hold, again, being properties (like here linearity) of the operator $\int$ itself rather than of the real plane. One such transposition from the real to the complex plane is not straightforward, namely, $\int_a^b f=\int_a^c f+\int_c^b f$, is not completely obvious at first sight, with $a$, $b$ and $c\in\mathbf{C}$, for instance if these points do not fall on the same line with $c$ in between. This is because, like for the case of derivability, the notion of path in complex calculus is an important one given the new degrees of freedom of the variables (from a line to a plane), and in general we need to specify which path is taken. We will see that, on the other hand, this is not always necessary for holomorphic functions, in a way which allows us to write identities such as Eq.~(1). The notation that specifies the path of integration is $\int_\mathcal{C}f(z)dz$ where $\mathcal{C}$ is the said path, that can be anything (a line, a curve, continuous or not, crossing itself, etc.)

For the case of a closed path, the notation $\oint$ is used. Paths that are not smooth can be broken into various paths on which the integration is computed piecewise, and the final result obtained by summing them up. When such a path is specified, we speak of a definite integral. Otherwise the integral is "indefinite".

Note that we could have computed $\int_0^i z\,dz$ as $\frac{1}{i}\int_0^i(iy)dy=\frac{1}{i^2}\int_0^1(iy)d(iy)=-1/2$, the integral as function of $iy$ being formally that already calculated in Eq.~(1). Here we have simply changed the variable, linking the integrals through the two paths as:

$$\tag{2} \int_\mathcal{C}f(z)\,dz=\int_\mathcal{D}f(g(\xi))g'(\xi)\,d\xi$$

with $\mathcal{D}$ the path followed by $\xi$. This is true provided $g$ is continuous and its derivative $g'$ is continuous on $\mathcal{D}$.

If the new path is a parametric function of a single variable~$t$, we can restore the lower/upper bounds notations. For example, $I=\int_\mathcal{C}\mathrm{Re}(z)\,dz$ where $\mathcal{C}$ is the straight line from $0$ to $1+2i$ can be computed through $\mathcal{D}=\{z\,,\> z(t)=t(1+2i)\,,\> 0\le t\le1\}$ and we arrive to $$I=\int_0^1\mathrm{Re}(z(t))z'(t)dt=\int_0^1t(1+2i)dt=i+1/2\,.$$

The complex integral is complex in general. A measure of its magnitude is given by its modulus, that can be easily bounded:

$$\tag{3} \left|\int_\mathcal{C}f(z)\,dz\right|\le ML$$

with $M$ such that $|f(z)|\le M$ for all $z\in\mathcal{C}$ and $L$ the length of $\mathcal{C}$. This is seen from the Riemann definition, every Riemann sum satisfying $$\left|\sum_{k=1}^N f(z_k)(z_k-z_{k-1})\right|\le\sum_{k=1}^N |f(z_k)|\times|(z_k-z_{k-1})|\le M\sum_{k=1}^N|(z_k-z_{k-1})|$$ with the latter sum being the length of $\mathcal{C}$ in the limit $N\rightarrow\infty$. This can be used to quickly check a result but more importantly is often used in a demonstration as an upper bound that can be brought to zero when taking the limit. For instance, $\int_\mathcal{C} z^2dz$ on the straight line between $0$ and~$1+i$ is $\le 2\sqrt2$.

Like has been seen many times before, it is not always necessary to go into the complex plane to get useful results. A trivial example is given by $\int_0^{\pi/2}e^{ax}\cos x\,dx=\frac{e^{a\pi/2}-a}{a^2+1}$, which follows right away from the real part of a complex exponential (from that we also get expressions for $\int x^ne^{ax}\cos x$ by derivation with respect to~$a$). A more involved example is provided by the integrals of rational functions, $P(x)/Q(x)$, which can always be obtained in closed form. If $\mathrm{deg}(P)<\mathrm{deg}(Q)$ (it can be brought to this form otherwise), there exists $c_{ij}\in\mathbf{C}$ such that:

$$\frac{P}{Q}=\sum_{i=1}^k\sum_{j=1}^{m_j}\frac{c_{ij}}{(x-\alpha_i)^j}$$

that can easily be computed through Eq.~(1). For instance, by the property just announced, there exists $A$, $B$, $C$ and~$D$ such that:

$$\frac1{1+x^4}=\frac{A}{x-(1+i)/\sqrt2}+\frac{B}{x-(1-i)/\sqrt2}+\frac{C}{x+(1-i)/\sqrt2}+\frac{D}{x+(1+i)/\sqrt2}$$

since the coefficients in the denominators are the roots of $1+x^4=0$ (no multiplicity). By multiplying both sides by the denominator of each expression of the rhs in turn and evaluating the left side (that cancels the pole), we find $A=\sqrt{2}(-1-i)/8$, $B=\sqrt{2}(-1+i)/8$, $C=\sqrt{2}(1-i)/8$ and $D=\sqrt{2}(1+i)/8$. The result is in term of complex logarithms, which might appear strange for the area of a real valued function, but this is correct since their imaginary parts cancel. This can be seen writting the result as: $\log(a+ib)=\frac{1}{2}\log(a^2+b^2)+i\arctan(b/a)$; beside, one can use $\arctan x+\arctan 1/x=\pm\pi/2$ depending on the sign of~$x$, to have the variable on the denominator; in this way, we arrive at:

$$\tag{4} \int\frac{dx}{x^4+1}=\frac{\sqrt 2}{8}\log\frac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}+ \frac{\sqrt2}{4}\left(\arctan(x\sqrt{2}+1)+\arctan(x\sqrt{2}-1)\right)\,.$$

We will see later in the course how to generalize this type of reasoning with the method of residues.

Whenever dealing with complex function, it is tempting to link them with the underlying real functions of real variables $f(z)=u(x,y)+iv(x,y)$. This gives, for the general case:

$$\int_\mathcal{C}f(z)\,dz=\int_\mathcal{C}(u+iv)(dx+idy)=\int_\mathcal{C}u\,dx-v\,dy+i\int_\mathcal{C} v\,dx+u\,dy\,.$$

Green's theorem provides such a link between the contour integral (on a line that closes on itself) and the surface integral through the surface~$\mathcal{R}$ enclosed by $\mathcal{C}$:

$$\tag{5} \oint_\mathcal{C}P\,dx+Q\,dy=\iint_\mathcal{R}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy\,.$$

The functions $P(x,y)$ and $Q(x,y)$ must be continuous with partial derivatives also continuous on $\mathcal{R}\cup\mathcal{C}$. We will see next lecture the properties of complex integrals for holomorphic functions.