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Let $$I=\int_{0}^{1}\int_{0}^{1} \left\{\frac{x}{y}\right\}\mathrm{d}x\mathrm{d}y.$$ When I tried computing the integral I seem to be getting a different answer to Wolramaplha, and can't find a similar integral anywhere on MSE or the internet.

Here's how I did it $$\int_{0}^{1}\left(\int_{0}^{1} \left\{\frac{x}{y}\right\}\mathrm{d}y\right)\mathrm{d}x$$, Lets evaluate, (Note: $0<x<1$) $$I_1=\int_{0}^{1} \left\{\frac{x}{y}\right\}\mathrm{d}y =\int_{1}^{\infty}\frac{xy-\lfloor xy \rfloor }{y^2}\mathrm{d}y=x(1-\gamma)$$

Therefore $I=\frac{(1-\gamma)}{2}$.

However Wolfram gives $I= 0.458868$.

Can someone help, or if my answer is wrong then provide a solution.

Sarthak Sahoo
  • 336

2 Answers2

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Let us do the things exactly: $$ \begin{aligned} I &= \int_{0}^{1}dx\int_{0}^{1} \left\{\frac xy\right\}\; dy \\ &\qquad\text{(Substitution: $y=x/z$)} \\ &= \int_0^1dx \int_x^\infty \left\{ z\right\}\;\frac x{z^2}\; dz \\ &= \int_0^1x\; dx \left(\ \int_x^1 \left\{ z\right\}\;\frac {dz}{z^2} + \int_1^\infty \left\{ z\right\}\;\frac {dz}{z^2} \ \right) \\ &= \int_0^1x\; dx \underbrace{\int_x^1 z\;\frac {dz}{z^2}}_{-\log x} \qquad +\qquad \underbrace{\int_0^1x\; dx}_{=1/2} \sum_{n\ge 1}\int_n^{n+1} \{z\}\;\frac {dz}{z^2} \\ &= \frac 14 + \frac 12\sum_{n\ge 1} \int_0^1 t\;\frac {dt}{(n+t)^2} \\ &= \frac 14 + \frac 12\sum_{n\ge 1} \left(\log\frac{n+1}n-\frac 1{n+1}\right) \\ &= \frac 14 + \frac 12\lim_{N\to\infty} \left(\log(N+1)-\frac 12-\frac 13-\dots-\frac1{N+1}\right) \\ &= \frac 14 + \frac 12\left(1-\gamma\right) \\ &= \frac 14(3-2\gamma) \\ &\approx 0.46139216754923\dots \end{aligned} $$

Computer checks:

Sage gives the numerical value for $(3-2\gamma)/4$:

sage: (3 - 2*euler_gamma.n())/4
0.461392167549234

Pari/gp has a better numerical integration, so i tried...

? \p 300
   realprecision = 308 significant digits (300 digits displayed)
? intnum(x=0, 1, intnum(y=0.00001, 1, frac(x/y)))
%20 = 0.4632302758838776079638791660602840407334814843635587641353594901119...

(Last result was manually truncated.)

? (3 - 2*Euler)/4
%22 = 0.4613921675492335696967439549587987844789203320300382005971163825575...
dan_fulea
  • 32,856
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Here is a different way of doing this. I assume the unstated fact, apparently obvious, that $\left\{\frac xy\right\}$ is the fractional part of $\frac xy$. Then \begin{aligned} I&=\int_0^1\int_0^1\left\{\frac xy\right\}dxdy=\int_0^1\left[ \int_x^1\frac xy dy+\int_0^x\left\{\frac xy\right\}dy\right]dx\\ &=-\int_0^1 x\ln x dx+\int_0^1\int_0^x\left\{\frac xy\right\}dydx \end{aligned} We now divide the interval from 0 to $x$ in the intervals $\frac x{n+1}<y<\frac xn$. Hence, in this interval, we have $n<\frac xy<n+1$. Therefore, in each such interval, $\frac xy=n+\left\{\frac xy\right\}$. Thus, we obtain \begin{aligned} I &=\frac14+\int_0^1\sum_{n=1}^\infty\int_{\frac x{n+1}}^{\frac xn}\left\{\frac xy\right\}dydx=\frac14+\int_0^1\sum_{n=1}^\infty\int_{\frac x{n+1}}^{\frac xn}\left[\frac xy-n\right]dydx\\ &=\frac 14+\int_0^1xdx\sum_{n=1}^\infty\left[\ln\left(\frac{n+1}n\right)-n\left(\frac1n-\frac1{n+1}\right)\right]\\ &=\frac 14+\frac12\sum_{n=1}^\infty\left[\ln\left(1+\frac1n\right)-\frac1{n+1}\right] \end{aligned} This is the same result obtained by @dan_fulea

minmax
  • 877