First slide

Evaluation of definite integrals

Question

For $m > 0, n > 0,$ let $I_{m, n} = \int_{0}^{1} x^{m} (\log x)^{n}$
$d x,$ then $I_{5, 5}$ is given by

Moderate

A

$- \frac{5!}{6^{5}}$
B

$- \frac{5!}{5^{5}}$
C

$- \frac{5!}{6^{6}}$
D

$\frac{5!}{6^{6}}$

Solution

Integrating by parts, we obtain
$I_{m, n} = {\frac{x^{m + 1} (\log x)^{n}}{m + 1}|}_{0}^{1} - \frac{n}{m + 1} \int_{0}^{1} x^{m} (\log x)^{n - 1} d x$
Since $lim_{x \to 0 +} x^{m} (\log x)^{n} = lim_{x \to 0 +} {(x^{m / n} \log x)}^{n} = 0$
So, $I_{m, n} = - \frac{n}{m + 1} I_{m, n - 1}$
Hence $I_{5, 5} = - \frac{5}{6} I_{5, 4}$
$\begin{array}{l} = - \frac{5}{6} (- \frac{4}{6} I_{5, 3}) \\ = (- 1)^{3} \frac{5.4.3}{6^{3}} I_{5, 2} = (- 1)^{4} \frac{5.4.3.2}{6^{4}} I_{5, 1} \\ I_{5, 1} = \int_{0}^{1} x^{5} \log x d x = {\frac{x^{6} \log x}{6}|}_{0}^{1} - \frac{1}{6} \int_{0}^{1} x^{6} \frac{1}{x} d x \\ = - \frac{1}{6^{2}} \end{array}$
So $I_{5, 5} = (- 1)^{5} \frac{5!}{6^{6}} = - \frac{5!}{6^{6}}$

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