symmetric double-integral on fractional part

Question

Let $\{\}$ denotes the fractional part function, does the following double-integral have a closed-form ?

$$\int_{0}^{1}\int_{0}^{1}\bigg\{\frac{1}{x}+\frac{1}{y}\bigg\}dx\,dy$$

Answer 1

It's not a full answer:

Using identity:

$$\left \{ z \right \}=\frac{1}{2}-\sum _{k=1}^{\infty } \frac{\sin (2 k \pi z)}{k \pi }$$

We can write (with CAS help):

$$\color{red}{\int_{0}^{1}\int_{0}^{1}\bigg\{\frac{1}{x}+\frac{1}{y}\bigg\}dx\,dy}=\\\int _0^1\int _0^1\left(\frac{1}{2}-\sum _{k=1}^{\infty } \frac{\sin \left(2 k \pi \left(\frac{1}{x}+\frac{1}{y}\right)\right)}{k \pi }\right)dydx=\\\frac{1}{2}-\sum _{k=1}^{\infty } \left(4 k \pi ^2 \text{Ci}(2 k \pi )+4 \cos (2 k \pi ) \text{Ci}(2 k \pi )+2 \pi \sin (2 k \pi )-\frac{\sin (4 k \pi )}{k \pi }+8 k \pi \text{Ci}(2 k \pi ) \text{Si}(2 k \pi )-4 \sin (2 k \pi ) \text{Si}(2 k \pi )\right)=\color{red}{\\\frac{3}{2}-2 \gamma +4 \pi \sum _{k=1}^{\infty } (2 k \text{Ci}(2 k \pi ) \text{Si}(2 k \pi )-k \pi \text{Ci}(2 k \pi ))}\approx0.495921$$

where: $\text{Ci}$ and $\text{Si}$ is cosine(sine) integral function.

Answer 2

This is not a solution as it does not provide a closed form but we derive a more compact form of the solution, in which the double integral over a fractional part is reduced to a single integral over a smooth integrand.

This is another illustration of the method described in my solution to Evaluation of $\int_{0}^{1}\int_{0}^{1}\{\frac{1}{\,x}\}\{\frac{1}{x\,y}\}dx\,dy\,$

Result

We show here that

$$i:=\int _0^1\int _0^1\{\frac{1}{x}+\frac{1}{y}\}\,dydx= 1-\gamma +\int_0^1 \psi ^{(1)}(\xi +1) \psi ^{(0)}(2-\xi ) \, d\xi\tag{1}$$

where $\{x\}$ is the fracional part of $x$, $\gamma$ is Euler gamma and $\psi ^{(n)}(x)$ is the polygamma function.

Here the integral over the polygamma functions is a $\simeq 17\%$ correction to $1-\gamma$. Numerically we have, respectively,

$$N(i)=0.42278433509846713 +0.07313656826103414=0.4959209033595013$$

Derivation

To begin with, we simplify the integral with the aim to reduce the fractional part of two variables to that of a single variable

Letting $x\to \frac{1}{r}$, $y\to \frac{1}{s}$, followed by $r\to u$, $s \to v-u$ leads to

$$i=\int _1^\infty \int _{1+u}^\infty \frac{1}{u^2 (v-u)^2}\{v\}\,dudv\tag{2}$$

Notice that the Jacobian determinant is $1$ and that, since $s\ge 1$, the v-integral has to start at $1+u$.

Next we split the integrals into separate integrals over integer regions, e.g.

$$\int_1^\infty f(u) \,du = \int_1^2 f(u) \,du +\int_2^3 f(u) \,du +... \\= \int_0^1 f(1+\xi) \,d\xi +\int_0^1 f(2+\xi) \,d\xi +... \\= \sum_{k=1}^\infty \int_0^1 f(k+\xi) \,d\xi = \int_0^1 (\sum_{k=1}^\infty f(k+\xi) )\,d\xi$$

in the last step we have optionally intechanged integration and summation.

In our case we let $u=k+\xi$, $v=m+\eta$ with $\{u\}=\xi$ and $\{v\}=\eta$ to obtain

$$i=\sum_{k=1}^\infty \int_0^1 \frac{\,d\xi}{(k+\xi)^2} \left(\int_{1+k+\xi}^{2+k} \frac{\,dv\{v\}}{(v-k-\xi)^2}\\+\sum_{m=2+k}^\infty \int_0^1 \frac{\,d\eta\; \eta}{(m-k-\xi +\eta)^2}\right)$$

The first term in the bracket gives after letting $v=(1+k+\xi)+\eta$, $\{v\}=\xi+\eta$ in the integral

$$\int_{0}^{1-\xi} \frac{\,d\eta(\xi+\eta)}{(v-k-\xi)^2}= \frac{1}{-2+\xi}+\xi +\log(2-\xi)$$

The second term in the bracket is with $m-k=n$

$$\sum_{n=2}^\infty \int_0^1 \frac{\,d\eta\; \eta}{(n-\xi +\eta)^2}\\=\sum_{n=2}^\infty (-\frac{1}{n-\xi +1}-\log (n-\xi )+\log (n-\xi +1))\\=\psi ^{(0)}(3-\xi )-\log (2-\xi )$$

Hence the bracket becomes

$$(\frac{1}{-2+\xi}+\xi +\log(2-\xi))+(\psi ^{(0)}(3-\xi )-\log (2-\xi ))\\=\frac{1}{-2+\xi}+\xi +\psi ^{(0)}(3-\xi ) $$

Notice that the log-term has dropped out.

Now we observe that the bracket simplifies further due to

$$\psi ^{(0)}(3-\xi )= H_{2-\xi}-\gamma$$

and

$$H_{2-\xi}-\frac{1}{-2+\xi}= H_{1-\xi}=\psi ^{(0)}(2-\xi )+\gamma$$

Since the bracket does not depend on $k$ the overall sum can be done

$$\sum_{k=1}^\infty \int_0^1 \frac{\,d\xi}{(k+\xi)^2}=\psi ^{(1)}(\xi +1)$$

and we are left with the integral

$$i=\int_0^1 \,d\xi \psi ^{(1)}(\xi +1)(\xi +\psi ^{(0)}(2-\xi )) $$

Finally, the relation

$$\int_0^1 \xi \psi ^{(1)}(\xi +1) \, d\xi=1-\gamma$$

completes the derivation.