$\lim_{R\mapsto\infty}\int_{-R}^{+R}\frac{1}{x+i}\text{d}x=?$

Question

I need help to solve the following integral:

$$\lim_{R\mapsto\infty}\int_{-R}^{+R}\frac{1}{x+i}\text{d}x.$$

None of the explicit formulas (with residues) of my course seems to apply directly. Can somebody help me please? Is the result simply $2\pi i$ by Cauchy’s integral formula? Thanks for your help!

Answer 1

\begin{align} \int_{-R}^R\frac 1{x+i}\mathrm dx &=\int_{-R}^R\frac{x-i}{x^2+1}\mathrm dx\\ &=\frac 12\int_{-R}^R\frac{2x}{x^2+1}\mathrm dx-i\int_{-R}^R\frac 1{x^2+1}\mathrm dx\\ &=\frac 12\left[\log(1+x^2)\right]_{-R}^R-i\left[\arctan(x)\right]_{-R}^R\\ &=-2i\arctan(R)\\ &\xrightarrow{R\to+\infty}-i\pi \end{align}

Answer 2

If you consider that $f(x)=x+i$ then the numerator is equal to $f'(x)$. The indefinite integral is than \begin{equation} \int_{}^{}\frac{1}{x+i}\text{d}x= log(x+i) + constant \end{equation} The indefinite integral is \begin{equation} \int_{-R}^{+R}\frac{1}{x+i}\text{d}x = log(R+i)-log(-R+i) \end{equation} for ($ Im(R) \geq -1$ or $Re(R)>0$) and ($Re(R) < 0$ or $Im(R)\leq 1$) and ($Re(R) \neq 0$ or $-1 < Im(R) < 0$ or $0 < Im(R) < 1$). If tou assume that R is positive you get $-2 i \arctan(R)$. You can then compute the limit which is $-i \pi$.