Another fractional part integral

Calculus Level 5

0 1 { 1 x } x 1 x d x = A γ + B \large \displaystyle\int _{ 0 }^{ 1 }{ \left\{ \dfrac { 1 }{ x } \right\} \dfrac { x }{ 1-x } \, dx=A\gamma } +B

The equation above holds true for integers A A and B B . Find A + B A+B .

Notations :


The answer is 1.

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.

3 solutions

Aditya Kumar
Apr 5, 2016

I = 0 1 { 1 x } x 1 x d x I=\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} } \frac { x }{ 1-x } dx

By Maclaurin series I'll write it as: I = k = 1 0 1 { 1 x } x k d x I=\sum _{ k=1 }^{ \infty }{ \int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} { x }^{ k }dx } }

Let J = 0 1 { 1 x } x k d x \displaystyle J=\int _{ 0 }^{ 1 }{ \left\{ \frac { 1 }{ x } \right\} { x }^{ k }dx }

Substitute: 1 x = x \frac { 1 }{ x } =x

J = i = 1 1 y i y k + 2 d y J=\sum _{ i=1 }^{ \infty }{ \int _{ 1 }^{ \infty }{ \frac { y-i }{ { y }^{ k+2 } } dy } }

On solving it, we get: J = 1 k ζ ( k + 1 ) k + 1 J=\frac { 1 }{ k } -\frac { \zeta \left( k+1 \right) }{ k+1 }

Now, coming back to our main sum: I = k = 1 ( 1 k ζ ( k + 1 ) k + 1 ) I=\sum _{ k=1 }^{ \infty }{ \left( \frac { 1 }{ k } -\frac { \zeta \left( k+1 \right) }{ k+1 } \right) }

I'll rearrange the terms as:

I = k = 1 ( 1 k 1 k + 1 ) k = 1 ζ ( k + 1 ) 1 k + 1 I=\sum _{ k=1 }^{ \infty }{ \left( \frac { 1 }{ k } -\frac { 1 }{ k+1 } \right) } -\sum _{ k=1 }^{ \infty }{ \frac { \zeta \left( k+1 \right) -1 }{ k+1 } }

Hence, I = γ \boxed{I=\gamma }

How do you know that k = 1 ζ ( k + 1 ) 1 k + 1 = 1 γ \displaystyle \sum _{ k=1 }^{ \infty }{ \frac { \zeta \left( k+1 \right) -1 }{ k+1 } } = 1 - \gamma ?

Pi Han Goh - 5 years, 1 month ago
Laurent Shorts
Apr 6, 2016

0 1 { 1 x } x 1 x d x = n = 1 1 n + 1 1 n ( 1 x n ) x 1 x d x = n = 1 1 n + 1 1 n ( n n 1 1 x ) d x \displaystyle\int_{0}^{1}{\left\{\frac{1}{x}\right\}}\frac{x}{1-x}\mathrm{d}x=\sum_{n=1}^{\infty}\int_{\frac{1}{n+1}}^{\frac{1}{n}}{\left(\frac{1}{x}-n\right)}\frac{x}{1-x}\mathrm{d}x=\sum_{n=1}^{\infty}\int_{\frac{1}{n+1}}^{\frac{1}{n}}{\left(n-\frac{n-1}{1-x}\right)\mathrm{d}x}

= 1 2 + n = 2 ( n ( 1 n 1 n + 1 ) + ( n 1 ) ( ln ( 1 1 n ) ln ( 1 1 n + 1 ) ) ) \displaystyle=\frac{1}{2}+\sum_{n=2}^{\infty}\left(n(\frac{1}{n}-\frac{1}{n+1})+(n-1)\big(\ln(1-\frac{1}{n})-\ln(1-\frac{1}{n+1})\big)\right)

= 1 2 + n = 2 ( 1 n + 1 + ( n 1 ) ( ln ( n 1 n ) ln ( n n + 1 ) ) ) \displaystyle=\frac{1}{2}+\sum_{n=2}^{\infty}\left(\frac{1}{n+1}+(n-1)\big(\ln(\frac{n-1}{n})-\ln(\frac{n}{n+1})\big)\right)

= 1 2 + n = 2 ( 1 n + 1 + ( n 1 ) ( ln ( n 1 ) 2 ln ( n ) + ln ( n + 1 ) ) ) \displaystyle=\frac{1}{2}+\sum_{n=2}^{\infty}\left(\frac{1}{n+1}+(n-1)\big(\ln(n-1)-2\ln(n)+\ln(n+1)\big)\right)

= 1 2 + lim N n = 2 N ( 1 n + 1 + ( n 1 ) ( ln ( n 1 ) 2 ln ( n ) + ln ( n + 1 ) ) ) \displaystyle=\frac{1}{2}+\lim\limits_{N \to \infty}\sum_{n=2}^{N}\left(\frac{1}{n+1}+(n-1)\big(\ln(n-1)-2\ln(n)+\ln(n+1)\big)\right)

= lim N n = 1 N ( 1 n + 1 ) + n = 2 N ( n 1 ) ln ( n 1 ) 2 n = 2 N ( n 1 ) ln ( n ) + n = 2 N ( n 1 ) ln ( n + 1 ) \displaystyle=\lim\limits_{N \to \infty}\sum_{n=1}^{N}\left(\frac{1}{n+1}\right)\,\,+\,\,\sum_{n=2}^{N}(n-1)\ln(n-1)-2\sum_{n=2}^N(n-1)\ln(n)+\sum_{n=2}^N(n-1)\ln(n+1)

= lim N n = 1 N ( 1 n + 1 ) + n = 1 N 1 n ln ( n ) 2 n = 2 N ( n 1 ) ln ( n ) + n = 3 N + 1 ( n 2 ) ln ( n ) \displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=1}^{N}\left(\frac{1}{n+1}\right)\,\,+\,\,\sum_{n=1}^{N-1}n\ln(n)-2\sum_{n=2}^N(n-1)\ln(n)+\sum_{n=3}^{N+1}(n-2)\ln(n)

= lim N n = 1 N ( 1 n + 1 ) + n = 3 N 1 n ln ( n ) 2 n = 3 N 1 ( n 1 ) ln ( n ) + n = 3 N 1 ( n 2 ) ln ( n ) N ln ( N ) + ( N 1 ) ln ( N + 1 ) \displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=1}^{N}\left(\frac{1}{n+1}\right)\,\,+\,\,\sum_{n=3}^{N-1}n\ln(n)-2\sum_{n=3}^{N-1}(n-1)\ln(n)+\sum_{n=3}^{N-1}(n-2)\ln(n)-N\ln(N)+(N-1)\ln(N+1)

= lim N n = 1 N ( 1 n + 1 ) N ln ( N ) + ( N 1 ) ln ( N + 1 ) \displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=1}^{N}\left(\frac{1}{n+1}\right)\,\,-\,N\ln(N)+(N-1)\ln(N+1)

= lim N n = 2 N + 1 ( 1 n ) + ln ( ( N + 1 ) N 1 N N ) \displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=2}^{N+1}\left(\frac{1}{n}\right)\,\,+\,\,\ln\left(\frac{(N+1)^{N-1}}{N^N}\right)

= lim N 1 + n = 1 N + 1 ( 1 n ) + ln ( ( N + 1 ) N N N ) ln ( N + 1 ) \displaystyle=\lim\limits_{{N \to \infty}}-1+\sum_{n=1}^{N+1}\left(\frac{1}{n}\right)\,\,+\,\,\ln\left(\frac{(N+1)^{N}}{N^N}\right)-\ln(N+1)

= lim N n = 1 N + 1 ( 1 n ) ln ( N + 1 ) + lim N ln ( ( 1 + 1 N ) N ) 1 \displaystyle=\lim\limits_{{N \to \infty}}\sum_{n=1}^{N+1}\left(\frac{1}{n}\right)\,-\,\ln(N+1)\,\,+\,\,\lim\limits_{{N \to \infty}}\ln\left(\big(1+\frac{1}{N}\big)^N\right)-1

= γ + ln ( e ) 1 = γ \displaystyle=\gamma\,\,+\,\,\ln\left(e\right)-1\,\,=\,\,\boxed{\gamma}

Wow!

nice solution! :)

Hamza A - 5 years, 2 months ago

Log in to reply

Lol your name is Hummus? XD

Sahil Silare - 3 years, 3 months ago
Shivam Sharma
Jun 14, 2017

See my solution...below...

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...