Integrating a function over a vertical line in the complex plane

Question

This is quite a simple question, and should hopefully have a simple answer.

How do I go about computing $$\int_{2-i\infty}^{2+i\infty} \frac{1}{t} \mathrm{d}t?$$

I've never seen such an integral before and don't know how to approach this.

Answer 1

Line integrals over complex functions are generally not path-independent. The function $\frac{1}{t}$ has a pole at $t=0$ so it is one of the many functions over which a line integral depends on the number of times the path being integrated over passes around the pole.

Answer 2

You should use complex analysis. In fact $$\int_{2-i\infty}^{2+i\infty} \frac{1}{t} \mathrm{d}t=\lim_{R\to\infty}\int_{2-iR}^{2+iR} \frac{1}{t} \mathrm{d}t$$also the function $f(t)=\dfrac{1}{t}$ is analytic and integrable throughout the integration path therefore$$\int_{2-iR}^{2+iR} \frac{1}{t} \mathrm{d}t=\ln \dfrac{2+iR}{2-iR}=\ln \dfrac{4-R^2+2iR}{4+R^2}$$also $$\lim_{R\to\infty}\dfrac{4-R^2+2iR}{4+R^2}=-1$$therefore $$\int_{2-i\infty}^{2+i\infty} \frac{1}{t} \mathrm{d}t=i\pi$$

Answer 3

You can also use residue theorem. Consider, for any $\xi>0$, the rectangle $\partial R_{\xi}$, where $$R_{\xi}=[-2,2]\times [-\xi,\xi] $$ and let $L,B,R,T$ be the left, bottom, right and top sides of $\partial R_{\xi}$ respectively, with anticlockwise parametrization. Our function has a simple pole in $t=0$ with residue $1$, and is holomorphic in $\mathbb{C}\setminus \left\{0\right\}$. Therefore, by residue theorem $$\int_{\partial R_{\xi}}\frac{1}{t}{dt}=2\pi i\qquad \forall \xi>0 $$ Now notice that $$\left|\int_{B}\frac{1}{t}dt\right| \leq \int_{-2-i\xi}^{-2+i\xi}\frac{1}{|t|}dt\leq \frac{4}{\xi}\to 0 $$ and the same goes for the top side. Thus $$2\pi i=\lim_{\xi\to \infty}\int_{\partial R_{\xi}}\frac{1}{t}dt=\lim_{\xi \to \infty}\int_L\frac{1}{t}dt+\lim_{\xi\to \infty}\int_R \frac{1}{t}dt $$ Also, $$\int_L\frac{1}{t}dt = \int_R\frac{1}{t}dt $$ This is because $\frac{1}{t}$ takes opposite values on symmetrical points with respect to the origin of the complex plane, but the parametrization is reversed, so the sign remains unchanged. Therefore $$i\pi = \lim_{\xi \to \infty}\int_{R}\frac{1}{dt}=\int_{2-i\infty}^{2+i\infty}\frac{1}{t}dt$$

Answer 4

If you define $\int_{2-i\infty}^{2+i\infty} f(t)dt $ as $\lim_{T \to \infty} \int_{2-iT}^{2+iT} f(t)dt $, then we can do the following:

Note that $z \mapsto {1 \over z}$ is analytic on the simply connected set $\operatorname{re} z > 0$, so if we integrate over the path $\gamma_T$ defined by $1 \to 2-iT \to 2+i T \to 1$ we get (using the standard branch of the $\log$) $\int_{\gamma_T} {1 \over z} dz = 0 = \log(2-iT)+ \int_{2-iT}^{2+iT} {1 \over t}dt + \log(2+i T)$.

Noting that $\operatorname{arg} (2+iT) = - \operatorname{arg} (2-iT)$, we obtain $\int_{2-iT}^{2+iT} {1 \over t}dt = \log|2 + iT|- \log|2-i T| + 2i\operatorname{arg} (2+iT) = \log | {2+iT \over 2-iT}| + 2i\operatorname{arg} (2+iT)$, and so, taking limits we get $\lim_{T \to \infty} 2i\operatorname{arg} (2+iT) = i\pi$.