Easiest!!

Calculus Level 4

The value of

$\int_0^1\left(\log\dfrac{1}{t}\right)^{1729}dt$

can be expressed as $N!.$ Find $N.$

The answer is 1729.

2 solutions

Sandeep Rathod
Nov 28, 2014

$log\frac{1}{x} = t , dt = -e^{-t}dx$

$\displaystyle\int_{\infty}^{0} -t^{n-1} e^{-t}dx$

$\displaystyle = \int_{0}^{\infty} t^{n-1} e^{-t}dx$

By definition of gamma function,

$= \displaystyle \int_{0}^{1}(log\frac{1}{x})^{n-1} = \int_{0}^{\infty} t^{n-1} e^{-t}dx = \Gamma(n) = 1729!$

You Can even do it without gamma Just using recursion You find That: $I_{n+1}=(n+1)*I_{n}$ and then You Can prove by induction That: $I_{n}=n!*I_{0}$ and $I_{0}=1$ so by putting 1729 we get the answer as 1729!

Oussama Boussif - 6 years, 6 months ago

The gamma function can be written as \Gamma(x) which results in LaTeX as

$\Gamma(x)$

Sharky Kesa - 6 years, 6 months ago

@Parth Lohomi you are talking about Mr.Sandeep Bhardwaj am i right

sandeep Rathod - 6 years, 6 months ago

Yes you got it

Parth Lohomi - 6 years, 6 months ago

I dont think this problem deserves level 5 ....its all easy if you know the gamma function...

Jayakumar Krishnan - 6 years, 6 months ago

hahaha..I got the answer and I am happy. thanks guys.

Sandeep Bhardwaj - 6 years, 6 months ago

Harsh Soni
Feb 15, 2015

Hey !! Nice Question

Easiest!!

The answer is 1729.

2 solutions

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