In summary, an Integral with Fractional Part is a mathematical expression that combines an integer and a fractional part. It is often used in calculus and other areas of mathematics to represent values between whole numbers and allows for more precise calculations. To evaluate it, the integer value is first found and added to the calculated fractional part. It can have a negative value depending on the values of the integer and fractional parts. Real-world applications include its use in engineering, physics, and economics to represent non-whole quantities such as distance or interest earned.
#1
Euge
Gold Member
MHB
POTW Director
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Evaluate the integral $$\int_0^1 x\left\{\frac{1}{x}\right\}\, dx$$ where ##\{\frac{1}{x}\}## denotes the fractional part of ##1/x##.
For ##y \ge 0##, we can rewrite the fractional part as ##\{y\} = y - \lfloor y \rfloor##. Thus
$$
\begin{align*}
\int_0^1 x \left\{ \frac{1}{x} \right\} dx &= \int_0^1 x \left( \frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor \right) dx \\
&= \int_0^1 dx - \int_0^1 x \left\lfloor \frac{1}{x} \right\rfloor dx \\
&= 1 - \int_0^1 x \left\lfloor \frac{1}{x} \right\rfloor dx
\end{align*}
$$
Setting ##y \equiv 1/x##, ##dy = -1/x^2 dx##, we have
$$
\begin{align*}
\int_0^1 x \left\lfloor \frac{1}{x} \right\rfloor dx &= \int_\infty^1 \frac{1}{y} \left\lfloor y \right\rfloor \left(-\frac{1}{y^2}\right) dy\\
&= \int_1^\infty \frac{1}{y^3} \left\lfloor y \right\rfloor dy
\end{align*}
$$
Since ##\left\lfloor y \right\rfloor## is piecewise constant, the integral can be written as the sum of integrals
$$
\begin{align*}
\int_1^\infty \frac{1}{y^3} \left\lfloor y \right\rfloor dy &= \int_1^2 \frac{1}{y^3} 1 dy +
\int_2^3 \frac{1}{y^3} 2 dy + \int_3^4 \frac{1}{y^3} 3 dy + \cdots \\
&= \sum_{n=1}^{\infty} \int_n^{n+1} \frac{1}{y^3} n dy \\
&= \sum_{n=1}^{\infty} n \int_n^{n+1} \frac{1}{y^3} dy \\
&= \sum_{n=1}^{\infty} n \left[ - \frac{1}{2y^2} \right]_n^{n+1} \\
&= \sum_{n=1}^{\infty} n \left[ \frac{1}{2n^2} - \frac{1}{2(n+1)^2}\right]
\end{align*}
$$
Here I have to admit that I cheated because I couldn't find a better way to write the sum, I thus couldn't find a simple expression for it. Mathematica told me it is ##\pi^2/12##, which is nice, but I still don't see how to rewrite the sum. Anyway, using this result, we finally find
$$
\begin{align*}
\int_0^1 x \left\{ \frac{1}{x} \right\} dx &= 1 - \frac{\pi^2}{12}
\end{align*}
$$
#4
julian
Gold Member
807
312
@DrClaude you do a shift in the summation variable ##n## followed by extending the sum:
##\sum_{n=1}^\infty n \frac{1}{2 (n+1)^2} = \sum_{n=2}^\infty (n-1) \frac{1}{2 n^2} = \sum_{n=1}^\infty (n-1) \frac{1}{2 n^2}##.