# Ask a about hard integral of $\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$

I want to evaluate the integral: $$I(a,b)=\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$$

Attempt:$$\frac{\partial ^2I}{\partial a\partial b}=4ab\int_{0}^{\infty}\frac{\log x}{(a^2+x^2)(b^2+x^2)}dx=\frac{4ab}{b^2-a^2}\int_{0}^{\infty}\log x\left(\frac{1}{a^2+x^2}-\frac{1}{b^2+x^2}\right)dx$$ $$=\frac{4ab}{b^2-a^2}\frac{\pi}{2}\left(\frac{\log a}{a}-\frac{\log b}{b}\right)=\frac{2\pi(b\log a-a\log b)}{b^2-a^2}$$ Then $$I(a,b)=2\pi\int_{0}^{b}\int_{0}^{a}\frac{(y\log x-x\log y)}{y^2-x^2}dxdy$$ But this integral very hard to solve,can anyone help me,thank you!

• Why not start by working small cases for $a$ and $b$, having a hard integral? – Number Sep 21 '18 at 10:26
• For case $I(a,a)=4a\pi \ln 2(\ln(a)-1)-\frac{a}{2}\pi^3$ – Number Sep 21 '18 at 10:30
• @Dahaka Thank you for your hint,but I still don't know how to do next. – JamesJ Sep 21 '18 at 12:21

## 2 Answers

This derivation is slightly different from Maxim's, because I'm not fluent in Meijer G. The beginning is the same, and assume $$0 Then

$$I(a,b)=b\,\int_0^\infty (\log{b} + \log{x} ) \log{(1+ \big(\frac{a/b}{x}\big)^2 ) }\log{(1+1/x^2)} dx$$

Let $$r=b/a \le 1$$ so that some series manipulations are permissible. The integral without the $$\log{x}$$ is easily performed in Mathematica:

$$\int_0^\infty \log{(1+ \big(r\,x\big)^{-2} ) }\log{(1+1/x^2)} dx = \pi\Big( (1+\frac{1}{r})(\text{arctanh(r)} - \log{(1-r^2)})-2\log{r}\Big)$$ Define $$J(s;r)=\int_0^\infty x^s \log{(1+ \big(r\,x\big)^{-2} ) }\log{(1+1/x^2)} dx \,.$$ The objective is to find $$\frac{d}{ds}J(s;r)\Big|_{s=0}$$ Within Mathematica J(s;r) can be found in terms of elementary functions and Gauss's hypergeometric $$F(a,b;c,x).$$

$$\frac{J(s;r)}{\pi}=\sec{(\frac{\pi s}{2})} \Big\{\!\frac{2\,r^2}{ (s\!+\!1)(s\!\!+3)}F(1,\!\frac{s+3}{2};\! \frac{s+5}{2}, r^2) - \frac{2\,r^{1-s}}{ (s\!+\!1)(s\!-\!1)}F(1,\!\frac{1-s}{2};\! \frac{3-s}{2}, r^2)$$ $$+ \frac{1}{s\!+\!1}\Big[ r^{-1-s}\log{(1-r^2)} + \log{(-1+1/r^2)}+\frac{2}{s+1} - \pi\,\tan{(\pi\,s/2)} \Big] \Big\}$$ Do the derivative and take $$s \to 0.$$ $$F(1,\frac{1}{2}; \frac{3}{2}, r^2)$$ and $$F(1,\frac{3}{2}; \frac{5}{2}, r^2)$$ evaluate to elementary functions. However the derivatives with respect to $$s$$ of the hypergeometrics do not. However, by using the series definition in terms of Pochhammer symbols, an easy calculation shows $$\frac{d}{ds} \frac{(3/2+s/2)_k}{(5/2+s/2)_k} \Big|_{s=0} = \frac{2k}{(2k+3)^2} \quad , \quad \frac{d}{ds} \frac{(1/2-s/2)_k}{(3/2-s/2)_k} \Big|_{s=0} = \frac{-2k}{(2k+1)^2}$$ In detail,

$$\frac{d}{ds} F(1,\frac{1-s}{2}; \frac{3-s}{2}, r^2) \Big|_{s=0} = -\sum_{k=0}^\infty\frac{2k}{(2k+1)^2} r^{2k} = -\sum_{k=0}^\infty\frac{2k+1 -1}{(2k+1)^2} r^{2k} =$$ $$=-\frac{\text{arctanh(r)}}{r} + \sum_{k=0}^\infty \frac{r^{2k}}{(2k+1)^2}= -\frac{\text{arctanh(r)}}{r} + \frac{1}{2r} \Big( \text{Li}_2(r) - \text{Li}_2(-r) \Big)$$ The series with the $$(2k+3)^2$$ in the denominator can be brought to this form with an index shift in the summation. Collect all the results and you finally get

$$\frac{I(a,b)}{\pi\,b}= \log{b} \Big( (1+\frac{1}{r})(2\,\text{arctanh}(r) + \log{(1-r^2)} ) -2\log{r} \Big)\, + \big(1-\frac{1}{r}\big) \big( \text{Li}_2(r) - \text{Li}_2(-r) \big)$$ $$-\Big(\frac{\pi^2}{2} + \log{(r^{-2}-1)} +2\,\big(1+\log{r}+\frac{1}{r} \big)\,\text{arctanh}(r) + \frac{1+\log{r}}{r} \, \log{(1-r^2)} \Big)$$

• $J(s;r)$ doesn't seem to be correct. I get $J(0; 1/2) = \pi (9 \ln 3 - 4 \ln 2)$ from the closed form, but the integral is $\pi ( 6 \ln 3 - 4 \ln 2)$. – Maxim Sep 21 '18 at 23:11
• @Maxim You are correct. There was a typo for the power of $r$ before the second hypergeometric. It read $r^{-1-s}$ and now reads $r^{1-s}.$ It has been checked numerically. Thanks. – skbmoore Sep 22 '18 at 15:44

One way is to replace $$\ln x$$ with $$x^p$$, then the integrand becomes a product of two linear Meijer G-functions after the change of variables $$t = 1/x^2$$. We obtain $$I(p) = \int_0^\infty x^p \ln \left(1 + \frac {a^2} {x^2} \right) \ln \left(1 + \frac {b^2} {x^2} \right) dx = \\ \frac 1 2 \int_0^\infty t^{(-3-p)/2} G_{2, 2}^{1, 2} \left( a^2 t \middle| {1, 1 \atop 1, 0} \right) G_{2, 2}^{1, 2} \left( b^2 t \middle| {1, 1 \atop 1, 0} \right) dt = \\ \frac {a^{1+p}} 2 G_{4, 4}^{3, 3} \left( \frac {b^2} {a^2} \middle| {1, 1, \frac {1+p} 2, \frac {3+p} 2 \atop 1, \frac {1+p} 2, \frac {1+p} 2, 0} \right),$$ which is expressible in terms of the Lerch transcendent. Then $$\int_0^\infty \ln x \ln \left(1 + \frac {a^2} {x^2} \right) \ln \left(1 + \frac {b^2} {x^2} \right) dx = I'(0) = \\ -\pi \left( \frac {a \omega (1-\omega)} 2 \Phi \!\left( \omega^2, 2, \frac 1 2 \right) + \frac {\pi^2 b} 2 - \\ (2b(1 - \ln b) - (a-b) \ln(1-\omega)) \ln \omega - (a+b) (\ln(a b) - 2) \ln(1+\omega) \right), \\ 0 < b < a, \quad\omega = \frac b a.$$