First slide

Evaluation of definite integrals

Question

The value of the integral $\int_{0}^{\infty} \frac{x \log x}{{(1 + x^{2})}^{2}} d x$ is

Moderate

A

7
B

0
C

$5 \log 13$
D

$2 \log 5$

Solution

$\begin{array}{l} \int_{0}^{\infty} \frac{x \log x}{{(1 + x^{2})}^{2}} d x \\ = \int_{0}^{1} \frac{x \log x}{{(1 + x^{2})}^{2}} d x + \int_{1}^{\infty} \frac{x \log x}{{(1 + x^{2})}^{2}} d x \end{array}$
Put $x = 1 / y$ in the second integral, so that $d x = (- 1 / y^{2}) d y$ . If
$x \to \infty$ then $y \to 0$ , and if $x = 1$ then $y = 1$ .
$\begin{array}{l} ∴ \int_{1}^{\infty} \frac{x \log x}{{(1 + x^{2})}^{2}} d x = \int_{1}^{0} \frac{1 \cdot \log y^{- 1}}{y {(1 + \frac{1}{y^{2}})}^{2}} (- \frac{1}{y^{2}}) d y \\ = - \int_{0}^{1} \frac{y \log y}{{(1 + y^{2})}^{2}} d y \\ \Rightarrow \int_{0}^{\infty} \frac{x \log x}{{(1 + x^{2})}^{2}} d x = \int_{0}^{1} \frac{x \log x}{{(1 + x^{2})}^{2}} d x - \int_{0}^{1} \frac{x \log x}{{(1 + x^{2})}^{2}} d x = 0 . \end{array}$

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