# Analytic/Numeric results to a list of integrals with fractional parts

I came across some integrals today that looked absolutely esoteric(from here). I tried to execute them in Mathematica but failed. Since we cannot always expect credibility from a random FB post, it could be a mere concoction. If so, how to debunk the claims?

1. $$\int\limits_0^1 {\int\limits_0^1 {{{\int\limits_0^1 {\left\{ {\frac{1}{{x + y + z}}} \right\}} }^n}{\mkern 1mu} dx{\mkern 1mu} dy{\mkern 1mu} dz} = \frac{{n!}}{{12}}\sum\limits_{k = 1}^\infty {\frac{{(k + 2)!}}{{(k + n)!}}(\zeta (k + 3) - 1) + \frac{{{2^{2 - n}} - {2^{1 - n}}}}{{n - 1}}} } + \frac{{n(1 - {2^{3 - n}})}}{{2(n - 1)(n - 2)}} + \frac{{{3^{3 - n}} - {2^{3 - n}}}}{{(n - 1)(n - 2)(n - 3)}}$$ for n$\geq$4, $\{\}$ denotes fractional part and $\zeta$ represents the Riemann zeta function

2. $$\int\limits_0^1 {\left\{ {{{\left\{ {\frac{{\Gamma (x)}}{{\sin (x)}}} \right\}}^\pi } - {{\left\{ {\frac{{\Gamma (1 - x)}}{{\sin (1 - x)}}} \right\}}^\pi }} \right\}} \ln (x)\ln (1 - x){\mkern 1mu} dx = 1 - \frac{{\zeta (2)}}{2}$$ where $\{\}$ is the fractional part, $\Gamma$ represents the Euler gamma function and $\zeta$ is the Riemann zeta function

3. $$\int\limits_0^1 {\left\{ {\frac{{{{( - 1)}^{\left[ {\frac{1}{x}} \right]}}}}{x}} \right\}} {\mkern 1mu} dx = 1 + \ln \left( {\frac{2}{\pi }} \right)$$

4. $$\int\limits_0^1 {\int\limits_0^1 {\int\limits_0^1 {\left\{ {\max \left( {\frac{1}{x},\frac{1}{y},\frac{1}{z}} \right)} \right\}} {\mkern 1mu} dx{\mkern 1mu} dy{\mkern 1mu} dz = 2\zeta (2) - 4\gamma } }$$

Below are extremely naive ways to check for Q3 and Q4 that I attempted. Clearly, there are some convergence issues.

Attempt of Q3:

In[6]:= NIntegrate[FractionalPart[(-1)^Floor[1/x]/x], {x, 0, 1}]

During evaluation of In[6]:= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In[6]:= NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near {x} = {0.0144815}. NIntegrate obtained -0.142899 and 0.006922707416829784 for the integral and error estimates.

Out[6]= -0.142899

In[7]:= 1 + Log[2/\[Pi]] // N

Out[7]= 0.548417


Attempt of Q4:

In[4]:= NIntegrate[
FractionalPart@Max[1/x, 1/y, 1/z], {x, 0, 1}, {y, 0, 1}, {z, 0, 1}]

During evaluation of In[4]:= NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

During evaluation of In[4]:= NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 0.5002708329538896 and 0.027491701668198695 for the integral and error estimates.

Out[4]= 0.500271

In[3]:= (2*Zeta[2] - 4*EulerGamma) // N

Out[3]= 0.981005


I will post a solution to the second integral: $$\int_0^1\!\!\mathrm{d}x\,\left\{\left\{\frac{\Gamma(x)}{\sin x}\right\}^{\pi}-\left\{\frac{\Gamma(1-x)}{\sin(1-x)}\right\}^{\pi}\right\}\ln x\ln(1-x).$$ Let $$I=\int_0^1\!\!\mathrm{d}x\,\left\{f(x)-f(1-x)\right\}\ln x\ln(1-x),\qquad(1)$$ where we have defined $$f(x)\equiv\left\{\frac{\Gamma(x)}{\sin x}\right\}^{\pi}.$$ Using the property $$\int_a^b\!\!\mathrm{d}x\,F(x)=\int_a^b\!\!\mathrm{d}x\,F(a+b-x),$$ we can write $$I=\int_0^1\!\!\mathrm{d}x\,\left\{f(1-x)-f(x)\right\}\ln(1-x)\ln x.\qquad(2)$$ Adding Eq. (1) and (2), and dividing by $2$ on both sides, we obtain $$I=\frac{1}{2}\int_0^1\!\!\mathrm{d}x\,\Big(\!\left\{f(x)-f(1-x)\right\}+\left\{f(1-x)-f(x)\right\}\!\Big)\ln x\ln(1-x).$$ We know that the fractional part $\{\cdot\}$ satisfies the property $$\{z\}+\{-z\}=\begin{cases}0~&\text{if z is an integer}\\1~&\text{otherwise}\end{cases},$$ hence the the terms enclosed within parenthesis equal $1$, and $$I=\frac{1}{2}\int_0^1\!\!\mathrm{d}x\,\ln x\ln(1-x).$$ Expanding $\ln(1-x)$ above in a Taylor series about $x=0$, we have \begin{align*}I&=-\frac{1}{2}\int_0^1\!\!\mathrm{d}x\,\ln x\left(x+\frac{x^2}{2}+\frac{x^3}{3}+\dotsi\right)\\&=-\frac{1}{2}\left(\int_0^1\!\!\mathrm{d}x\,x\ln x+\frac{1}{2}\int_0^1\!\!\mathrm{d}x\,x^2\ln x+\frac{1}{3}\int_0^1\!\!\mathrm{d}x\,x^3\ln x+\dotsi\right)\\&=\frac{1}{2}\left(\frac{1}{2^2}+\frac{1}{2\cdot3^2}+\frac{1}{3\cdot4^2}+\dotsi\right).\qquad(3)\end{align*} In order to evaluate the integrals preceeding (3), we can use an interesting trick, viz., \begin{align*}\int_0^1\!\!\mathrm{d}x\,x^k\ln x&=\int_0^1\!\!\mathrm{d}x\left(\frac{\partial}{\partial a}x^a\right)_{\!a=k}=\left[\frac{\mathrm{d}}{\mathrm{d}a}\int_0^1\!\!\mathrm{d}x\,x^a\right]_{a=k}\\&=\left[\frac{\mathrm{d}}{\mathrm{d}a}\left(\frac{1}{a+1}\right)\right]_{a=k}=-\frac{1}{(k+1)^2}.\end{align*} Now, returning to the evaluation of (3) \begin{align*}I&=\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k(k+1)^2}=\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}-\frac{1}{(k+1)^2}\right)\\&=\frac{1}{2}\sum_{k=1}^{\infty}\left(\frac{1}{k}-\frac{1}{k+1}\right)-\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{(k+1)^2}\qquad(4)\\&=\frac{1}{2}-\frac{1}{2}(\zeta(2)-1)=1-\frac{\zeta(2)}{2}\\&=1-\frac{\pi^2}{12}.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{Q.E.D}\end{align*} Note, in Eq. (4) the first sum is a telescoping series, while the second sum is simply the Riemann zeta function of $2$, minus $1$.

Cheers!

• Makes me wonder if this Q&A should be on math.stackexchange.com. Where's the code showing how to evaluate the integral in Mathematica? Apr 10 '18 at 16:57

Q3 right hand side seems to be wrong:

f[n_] := Integrate[ FractionalPart[(-1)^Floor[1/x]/x], {x, 1/(n + 1), 1/n}]

FullSimplify[Sum[f[k], {k, 1, 100}]]

2126095643991513237687041380525277645295683/\ 7041757898200960193617914702466542659236800 - 194 Log[2] + 8 Log[3] + Log[2448382020236497666691277794111678553551478029775138509]

N[%]

-0.144705

FullSimplify[Sum[f[k], {k, 1, 1000}]]

2183953315830054714446779905044622572049495527462187031174191401152135\ 5698579361689143814264535541146502973594611338778997749755767809986443\ 0241934714638791259613988070137105544798560120867369724977430318669179\ 8613579320063766884598344228871637429247809010874110522953324937108540\ 2956774115434480895448779975401244940757039022404682890424661215840634\ 4670802939679011936513569250248101083268116349105489153587027879671549\ 1961090635733/ 71288652746650930531663841557142729206683588618858930404520019911543\ 2408758111149947644415191387158691171781701957525651298026406762100925\ 1465871004305131072686268143200196609974862745937188343705015434452523\ 7397452989631456749821282369562328237940110688092623177088619795407912\ 4775455804932647573782992335275179673524804246363805113703433121478174\ 6850878453485678021888075373249921995672056932029099390891687487672697\ 950931603520000 - 1988 Log[2] + 8 Log[3] + 4 Log[23] + Log[35870834647263637116458899707597523387989181660361424756820427969\ 3763914030996514681350255254583299596078296626803024781012146366427697\ 84609386929780987791033225] + 2 Log[164646606772823990543027983091426282825597063348423869493686896\ 1270303723508041575984369549297782268253971936882295461857930061431060\ 2828935775683539562170207160751788624421695930927459307369720254613378\ 466544184313]

N[%]

-0.14473

So it appears to converge to -0.1447... which is close to the NIntegrate` answer.

I see little reason to trust the other claims...