A calculus problem by Md Zuhair

Calculus Level 5

0 1 { 1 2017 x 2017 } d x \large{\displaystyle{\int^{1}_{0} \left \{\dfrac{1}{2017 \space \sqrt[2017]{x}} \right\} \ dx}}

If the closed form of the integral above can be represented as 1 b ζ ( c ) d e \dfrac{1}{b} - \dfrac{\zeta (c)}{d^e} , where a a and b b are co-prime integers, then evaluate b + c + d + e b+c+d+e .

Notations

Inspiration Tapas Mazumdar . Try this first.


The answer is 8067.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Mark Hennings
May 28, 2017

As shown in the inspiration question, we have 0 1 { 1 x n } d x = N 1 { t } t N 1 d t = N N 1 ζ ( N ) \int_0^1 \left\{\frac{1}{\sqrt[n]{x}}\right\}\,dx \; = \; N\int_1^\infty \{t\} t^{-N-1}\,dt \;= \; \tfrac{N}{N-1} - \zeta(N) In this problem, the substitutions y = N x N y = N\sqrt[N]{x} and t = y 1 t = y^{-1} give 0 1 { 1 N x N } d x = 1 N N 1 0 N { y 1 } y N 1 d y = 1 N N 1 1 N { t } t N 1 d t \int_0^1 \left\{ \frac{1}{N \sqrt[N]{x}}\right\}\,dx \; = \; \frac{1}{N^{N-1}} \int_0^N \left\{y^{-1}\right\}y^{N-1}\,dy \; = \; \frac{1}{N^{N-1}} \int_{\frac{1}{N}}^\infty \{t\} t^{-N-1}\,dt and hence the integral is equal to 1 N N 1 ( 1 N 1 { t } t N 1 d t + 1 N 1 ζ ( N ) N ) = 1 N N 1 ( 1 N 1 t N d t + 1 N 1 ζ ( N ) N ) \frac{1}{N^{N-1}}\left(\int_{\frac{1}{N}}^1 \{t\}t^{-N-1}\,dt + \frac{1}{N-1} - \frac{\zeta(N)}{N}\right) \; = \; \frac{1}{N^{N-1}}\left(\int_{\frac{1}{N}}^1 t^{-N}\,dt + \frac{1}{N-1} - \frac{\zeta(N)}{N}\right) which evaluates as 1 N 1 ζ ( N ) N N \frac{1}{N-1} - \frac{\zeta(N)}{N^N} With N = 2017 N=2017 , this makes the answer 2016 + 2017 + 2017 + 2017 = 8067 2016 + 2017 + 2017 + 2017 = \boxed{8067} .

@Mark Hennings sir . Can we both make a wiki on fractional integration?

Md Zuhair - 4 years ago

Log in to reply

Log in to reply

Oh . It does. Thank you Sir :)

Md Zuhair - 4 years ago
First Last
May 28, 2017

0 1 { 1 2017 x 1 2017 } d x = 0 1 1 2017 x 1 2017 d x 0 1 1 2017 x 1 2017 d x = \displaystyle\int_0^1\bigg\{\frac1{2017x^\frac1{2017}}\bigg\}dx = \int_0^1\frac1{2017x^\frac1{2017}}dx - \int_0^1\lfloor\frac1{2017x^\frac1{2017}}\rfloor dx =

The below summation is created from seeing where the floor function increments, with the integral interval preserved.

1 2016 n = 1 ( n 2017 ) 2017 ( ( n + 1 ) 2017 ) 2017 n d x = \displaystyle\frac1{2016}-\sum_{n=1}^\infty\int_{(n*2017)^{-2017}}^{((n+1)*2017)^{-2017}}ndx =

1 2016 1 201 7 2017 n = 1 n ( 1 ( n + 1 ) 2017 1 n 2017 ) = 1 2016 ζ ( 2017 ) 201 7 2017 \displaystyle\frac1{2016}-\frac1{2017^{2017}}\sum_{n=1}^\infty n*\bigg(\frac1{(n+1)^{2017}}-\frac1{n^{2017}}\bigg) = \boxed{\frac1{2016}-\frac{\zeta(2017)}{2017^{2017}}}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...