Zeros are heros

Calculus Level 5

Find the number of trailing zeros in the answer of the integral

$\large{\displaystyle \int_{0}^{1} (ln \frac{1}{x})^{85}.dx}$

The answer is 20.

2 solutions

Tanishq Varshney
Jun 7, 2015

Dont know the proof

$\huge{\displaystyle \int^{1}_{0} (ln \frac{1}{x})^{n}.dx=n!}$

$\large{\forall~n>-1}$

And to find the number of trailing zeros in $n!$

$[\frac{n}{5}]+[\frac{n}{25}]+[\frac{n}{125}]+[\frac{n}{625}]+......$

where $[...]$ is floor function.

here $n=85$ , number of trailing zeros $=17+3=20$

Moderator note:

You failed to show the most important part of the proof: How do you know that $\displaystyle \int_0^1 \left( \ln \frac 1x \right)^n \, dx = \frac 1n$ ?

shouldn't $log \frac {1}{x}$ be $ln \frac {1}{x}$ ? Because $\large \int _0^1\left(log\frac{1}{x}\right)^ndx\: = \frac{n!}{ln^n\left(10\right)}$ .

Efren Medallo - 6 years ago

edited thanks

Tanishq Varshney - 6 years ago

James Wilson
Jan 11, 2018

If you simply make the substitution $u=\ln{\frac{1}{x}}$ , then the integral becomes $\int_{\ln{\frac{1}{0+}}}^0 u^{85}(-x)du=\int_0^\infty u^{85}e^{-u}du$ . This is the gamma function evaluated at 86, which is $85!$ . The final step then is to figure out how many times you can divide $85!$ evenly by $5$ .

Zeros are heros

The answer is 20.

2 solutions

Moderator note:

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