### Author Topic: Term Test 2 sample P4  (Read 2225 times)

#### Victor Ivrii

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##### Term Test 2 sample P4
« on: October 29, 2018, 05:19:56 AM »
Calculate an improper integral
$$I=\int_0^\infty \frac{\ln^2(x)\,dx}{(x^2+1)}.$$

Hint: (a) Calculate
$$J_{R,\varepsilon} = \int_{\Gamma_{R,\varepsilon}} f(z)\,dz, \qquad f(z):=\frac{\log^2(z)}{(z^2+1)}$$
where we have chosen the branch  of $\log(z)$ 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 $\int_{\gamma_R} f(z)\,dz\to 0$ as $R\to \infty$ and $\int_{\gamma_\varepsilon} f(z)\,dz\to 0$ 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.
\label{4-1}

(c) Express both integrals using $I$.

« Last Edit: October 29, 2018, 05:24:33 AM by Victor Ivrii »

#### Heng Kan

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##### Re: Term Test 2 sample P4
« Reply #1 on: November 11, 2018, 09:28:46 PM »
As this question is a bit troublesome, there are two scanned pictures.  The final answer is π^3/8, which is shown in the last line of the second picture.

#### Victor Ivrii

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##### Re: Term Test 2 sample P4
« Reply #2 on: November 12, 2018, 04:14:18 AM »
Difficult to read. Completely insufficient space between lines.

On test and exam grader could just miss correct elements of the solution.

#### Heng Kan

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##### Re: Term Test 2 sample P4
« Reply #3 on: November 12, 2018, 01:30:48 PM »
I think it is OK  this time. There are 4 pictures. The first one is for part(a), the second and third one are for part(b), and the last one is for part (c)

#### Victor Ivrii

Estimate over large semi-circle contains fixable errors (I would show on the typed solution) and is over-complicated: The integrand does not exceed $(R-1)^{-2}(|\ln R +5)^2$ (rough estimate) and the  integral does not exceed this multiplied by $\pi R$, so it tends to $0$ as $R\to \infty$
Estimate over large semi-circle also contains errors and is over-complicated: Integrand does not exceed $2|\ln \varepsilon|^2$ nd the  integral does not exceed this multiplied by $\pi \varepsilon$, , so it tends to $0$ as $\varepsilon\to 0$.