Rectangles in use of complex Analysis

Question

(a) I think this is by the use of the estimation lemma. So I wrote But I don't know how to get the 1/n!?

(b) I did it again by the estimation lemma and i got the required result.

(c) I don't know exactly but i assume we take a parametrisation of Rn counterclockwise and that Rn is the union of the sides. But i am stuck on how to proceed.

Answer 1

For Part c), we write

$$\begin{align} \oint_{R_n}\frac{e^{-z}}{z+1}\,dz&=\int_0^n \frac{e^{-(x-in)}}{(x+1)-in}\,dx\\\\ &+\int_{-n}^n\frac{e^{-(n+iy)}}{(n+1)+iy}\,i\,dy\\\\ &-\int_0^n\frac{e^{-(x+in)}}{(x+1)+in}\,dx\\\\ &-\int_{-n}^n\frac{e^{-iy}}{1+iy}\,i\,dy \tag 1 \end{align}$$

Since $\frac{e^{-z}}{1+z}$ is analytic in and on $R_n$, Cauchy's Integral Theorem guarantees that the integral on the left-hand side of $(1)$ vanishes. Alongside this, note that as $n\to \infty$, the first, second, and third integrals on the right-hand side of $(1)$ tend to $0$.

Therefore, we have

$$\begin{align} 0&=\int_{-\infty}^\infty\frac{e^{-iy}}{1+iy}\,dy\\\\ &=\int_{-\infty}^\infty\frac{(1-iy)e^{-iy}}{1+y^2}\,dy\\\\ &=\int_{-\infty}^\infty\frac{\cos(y)-y\sin(y)}{1+y^2}\,dy-i\int_{-\infty}^\infty\frac{y\cos(y)+\sin(y)}{1+y^2}\,dy \\\\ &=\int_{-\infty}^\infty\frac{\cos(y)-y\sin(y)}{1+y^2}\,dy \tag 2 \end{align}$$

since the imaginary part is the integral of an odd integrand integrated over anti-symmetric limits, and hence $0$.

Finally, rearranging $(2)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty\frac{\cos(y)}{1+y^2}\,dy=\int_{-\infty}^\infty\frac{y\sin(y)}{1+y^2}\,dy}\tag 3$$

as was to be shown!

Interestingly, the integrals in $(3)$ can be evaluated in closed form for which

$$\int_{-\infty}^\infty\frac{\cos(y)}{1+y^2}\,dy=\int_{-\infty}^\infty\frac{y\sin(y)}{1+y^2}\,dy =\pi e^{-1}$$

Answer 2

For (a), in $\Gamma_n^B$, $z=t-in$ and hence $dz=dt$, $0\le t\le n$. So \begin{eqnarray} &&\bigg|\int_{\Gamma_n^B}\frac{e^{-z}}{z+1}dz\bigg|\\ &=&\bigg|\int_0^n\frac{e^{-t+in}}{t-in+1}dt\bigg|\\ &\le&\int_0^n\frac{e^{-t}}{\sqrt{(t+1)^2+n^2}}dt\\ &\le&\frac{1}{n}\int_0^ne^{-t}dt\\ &\le&\frac{1}{n}. \end{eqnarray} You can do the same trick for other estimates.