MAT334-2018F > Term Test 2

TT2B Problem 4

(1/2) > >>

Victor Ivrii:
Calculate an improper integral
$$
I=\int_0^\infty \frac{\ln(x)\sqrt{x}\,dx}{(x^2+1)}.
$$

Hint:
 
(a) Calculate
$$
J_{R,\varepsilon} = \int_{\Gamma_{R,\varepsilon}} f(z)\,dz, \qquad f(z)=\frac{\sqrt{z}\log(z)}{(z^2+1)}
$$
where we have chosen the branches of $\log(z)$ and $\sqrt{z}$ such that they are analytic on the upper half-plane $\{z\colon \Im z>0\}$ and is real-valued for $z=x>0$. $\Gamma_{R,\varepsilon}$ is the contour on the figure below:

(b)  Prove that $\int_{\gamma_R}  \frac{\sqrt{z}\log(z)\,dz}{(z^2+1)}\to 0$ as $R\to \infty$ and $\int_{\gamma_\varepsilon}  \frac{\sqrt{z}\log(z)\,dz}{(z^2+1)}\to 0$ as $\varepsilon\to 0^+0$ where $\gamma_R$ and $\gamma_\varepsilon$ are large and small semi-circles on the picture. This will give you a value of
$$
\int_{-\infty}^0 f(z)\,dz + \int_0^{\infty} f(z)\,dz.
\tag{*}
$$
 
(c) Express both integrals using $I$.

Zixuan Miao:
See attached.

Wrong calculation at that part

Zixuan Miao:
.

Victor Ivrii:
Zixuan,
please correct

hanyu Qi:
(a)

By residue thrm, $$ \int_{\gamma_{R,\epsilon}} f(z) \text{d}z = 2\pi i Res(f(z),i) = \lim_{z \rightarrow i} (z-i) f(z) = \frac{\ln i \sqrt i}{2i} = \frac{\pi \sqrt i }{4} = \frac{ \sqrt 2 \pi}{8} + i \frac{\sqrt 2 \pi}{8} $$

Since $$ \sqrt i = \frac{\sqrt 2 }{2} + \frac{\sqrt 2 i}{2} $$

$$ \ln i = \ln |i| + i arg(i) = i \frac{ \pi }{2} $$

Navigation

[0] Message Index

[#] Next page

Go to full version