Integral with fractional part

Calculus (Integrals, Series)
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Hamza Mahmood
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Integral with fractional part

#1

Post by Hamza Mahmood »

Let $n \in \mathbb{N} \setminus \{1, 2\}$. Prove that:

$$\int_1^\infty \frac{\{x\}}{x^n}\, dx = \frac{1}{n-2} - \frac{\zeta(n-1)}{n-1}$$
r9m
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Re: Integral with fractional part

#2

Post by r9m »

\begin{align*}\int_1^{\infty} \frac{\{x\}}{x^n}\,dx &= \sum\limits_{k=1}^{\infty}\int_k^{k+1} \frac{\{x\}}{x^n}\,dx\\
&= \sum\limits_{k=1}^{\infty} \int_k^{k+1} \frac{x - k}{x^n}\,dx\\
&= \sum\limits_{k=1}^{\infty} \left[\frac{x^{2-n}}{2-n} - k\frac{x^{1-n}}{1-n}\right]_k^{k+1}\\
&= \sum\limits_{k=1}^{\infty} \left(\frac{1}{n-2}\left(\frac{1}{k^{n-2}} - \frac{1}{(k+1)^{n-2}}\right) - \frac{1}{n-1}\left(\frac{1}{k^{n-2}} - \frac{k+1-1}{(k+1)^{n-1}}\right)\right)\\&= \frac{1}{n-2} - \frac{1}{n-1}\sum\limits_{k=1}^{\infty}\frac{1}{k^{n-1}}\\&=\frac{1}{n-2} - \frac{\zeta(n-1)}{n-1}\end{align*}
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Tolaso J Kos
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Re: Integral with fractional part

#3

Post by Tolaso J Kos »

Hamza Mahmood wrote:Let $n \in \mathbb{N} \setminus \{1, 2\}$. Prove that:

$$\int_1^\infty \frac{\{x\}}{x^n}\, dx = \frac{1}{n-2} - \frac{\zeta(n-1)}{n-1}$$
Hello Hamza,

recalling the fact that $\left \{ x \right \}= x - \left \lfloor x \right \rfloor$ where $\left \lfloor \cdot \right \rfloor$ is the floor function , then for our integral we have successively:

\begin{align*}
\int_{1}^{\infty}\frac{\left \{ x \right \}}{x^n}\, {\rm d}x &=\int_{1}^{\infty} \frac{x- \left \lfloor x \right \rfloor}{x^n}\, {\rm d}x \\
&= \int_{1}^{\infty}\frac{{\rm d}x}{x^{n-1}} - \int_{1}^{\infty}\frac{\left \lfloor x \right \rfloor}{x^n}\, {\rm d}x\\
&= \frac{1}{n-2} - \int_{1}^{\infty}\frac{\left \lfloor x \right \rfloor}{x^n}\, {\rm d}x\\
&= \frac{1}{n-2} - \sum_{k=1}^{\infty}k \int_{k}^{k+1}\frac{{\rm d}x}{x^n}\\
&= \frac{1}{n-2} + \frac{1}{n-1}\sum_{k=1}^{\infty} k \left [ \frac{1}{\left ( k+1 \right )^{n-1}}- \frac{1}{k^{n-1}}\right ] \\
&=\frac{1}{n-2} - \frac{1}{n-1} \sum_{k=1}^{\infty}\frac{1}{k^{n-2}} + \frac{1}{n-1}\sum_{k=1}^{\infty}\frac{k}{(k+1)^{n-1}}\\
&=\frac{1}{n-2} - \frac{1}{n-1} \zeta(n-2) + \frac{1}{n-1}\sum_{k=1}^{\infty} \frac{k+{\color{red}{1 -1}}}{\left ( k+1 \right )^{n-1}} \\
&=\frac{1}{n-2} - \frac{1}{n-1} \zeta(n-2) + \frac{1}{n-1} \sum_{k=1}^{\infty}\left [ \frac{1}{\left ( k+1 \right )^{n-2}} - \frac{1}{\left ( k+1 \right )^{n-1}} \right ] \\
&= \frac{1}{n-2} - \frac{1}{n-1} \zeta(n-2) + \frac{1}{n-1}\left[\zeta(n-2) -1\right] + \frac{1}{n-1} \left[1-\zeta(n-1)\right] \\
&=\frac{1}{n-2}\cancel{- \frac{\zeta(n-2)}{n-1} + \frac{\zeta(n-2)}{n-1}} \bcancel{- \frac{1}{n-1} + \frac{1}{n-1}} -\frac{\zeta(n-1)}{n-1} \\
&=\frac{1}{n-2} - \frac{\zeta(n-1)}{n-1}
\end{align*}

that is the result you stated. :) Here $\zeta$ stands for the Riemann Zeta function.

Oops, I came second. Never mind. I leave the post just because I typed all this.
Imagination is much more important than knowledge.
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