MAT334-2018F > Reading Week Bonus--sample problems for TT2

Term Test 2 sample P4

(1/1)

**Victor Ivrii**:

$\renewcommand{\Re}{\operatorname{Re}}

\renewcommand{\Im}{\operatorname{Im}}$

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

\begin{equation}

\int_{-\infty}^0 f(z)\,dz + \int_0^{\infty} f(z)\,dz.

\label{4-1}

\end{equation}

(c) Express both integrals using $I$.

**Heng Kan**:

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**:

Difficult to read. Completely insufficient space between lines.

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

**Heng Kan**:

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 $\varepsilon\to 0$.

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