Author Topic: TT2B Problem 4  (Read 869 times)

Victor Ivrii

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TT2B Problem 4
« on: November 24, 2018, 05:25:15 AM »
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 $I$.
« Last Edit: December 02, 2018, 12:11:13 PM by Victor Ivrii »

Zixuan Miao

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Re: TT2B Problem 4
« Reply #1 on: November 24, 2018, 09:31:44 AM »
See attached.

Wrong calculation at that part
« Last Edit: November 29, 2018, 09:18:09 AM by Victor Ivrii »

Zixuan Miao

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Re: TT2B Problem 4
« Reply #2 on: November 24, 2018, 10:03:59 AM »
.

Victor Ivrii

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Re: TT2B Problem 4
« Reply #3 on: November 29, 2018, 09:21:13 AM »
Zixuan,
please correct

hanyu Qi

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Re: TT2B Problem 4
« Reply #4 on: December 01, 2018, 05:10:08 PM »
(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 $$ \sqrt i = \frac{\sqrt 2 }{2} + \frac{\sqrt 2 i}{2} $$

$$ \ln i = \ln |i| + i arg(i) = i \frac{ \pi }{2} $$
« Last Edit: December 01, 2018, 05:31:02 PM by Alex Qi »

Victor Ivrii

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Re: TT2B Problem 4
« Reply #5 on: December 02, 2018, 12:08:18 PM »
After Alex Qi correction from Zixuan post it follows $\newcommand{\Res}{\operatorname{Res}}$

(a) As $R>1$ there is just one singularity inside $\Gamma_{R,\varepsilon}$, namely a simple pole at $z=i$. The residue is
$\Res (\frac{\sqrt{z}\log(z)}{(z^2+1)}, i)= \frac{\sqrt{z}\log(i)}{{2z}}|_{z=i}= \frac{\pi i/2 }{2i}e^{i\pi/4}$ due to the branch selection and therefore $J= 2\pi i \times \frac{\pi i/2 }{2i}e^{i\pi/4}= \frac{\pi^2}{2} e^{3i\pi/4}=
\frac{\pi^2}{2\sqrt{2}} (-1-i)$.

(b)
$$
\int_{\gamma_R}  \frac{\sqrt{z}\log(z)\,dz}{(z^2+1)}|\le \frac{  \pi R \sqrt{R}(\ln(R)+\pi)}{(R-1)^2}\to 0\qquad \text{as }\ \ R\to \infty
$$
and
$$
\int_{\gamma_\varepsilon}  \frac{\sqrt{z}\log(z)\,dz}{(z^2+1)}|\le \frac{ \pi \varepsilon \sqrt{\varepsilon} (|\log \varepsilon|+\pi)}{(1-\varepsilon)^2}\to 0\qquad \text{as  }\ \ \varepsilon \to 0
$$
(c) In (*) the second integral is $I$ and the first one is
$$
\int_{-\infty}^0 \frac{e^{i\pi/2}\sqrt{|x|}(\ln |x|+\pi i)\, dx}{(x^2+1)} =
i\int_{\infty}^0 \frac{\sqrt{|x|} (\ln |x|+\pi i) dx}{(x^2+1)}=
i I -\pi K
$$
with
$$
K=\int_0^\infty\frac{\sqrt{|x|}  dx}{(x^2+1)}
$$
after change of variables. Thus
$$
(i+1)I -\pi K= \frac{\pi^2}{2\sqrt{2}} (-1+i).
$$
Since both $I,K$ are real
$$
I =\frac{\pi^2}{2\sqrt{2}}.
$$
As an added value we get $I-\pi K =-\frac{\pi^2}{2\sqrt{2}}$ and $K=\frac{\pi}{\sqrt{2}}$.
« Last Edit: December 02, 2018, 12:10:37 PM by Victor Ivrii »