5
$\begingroup$

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!

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

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

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

$\endgroup$
  • $\begingroup$ $J(s;r)$ 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 $\pi ( 6 \ln 3 - 4 \ln 2)$. $\endgroup$ – Maxim Sep 21 '18 at 23:11
  • $\begingroup$ @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. $\endgroup$ – skbmoore Sep 22 '18 at 15:44
5
$\begingroup$

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 $$\int_0^\infty \ln x \ln \left(1 + \frac {a^2} {x^2} \right) \ln \left(1 + \frac {b^2} {x^2} \right) dx = I'(0) = \\ -\pi \left( \frac {a \omega (1-\omega)} 2 \Phi \!\left( \omega^2, 2, \frac 1 2 \right) + \frac {\pi^2 b} 2 - \\ (2b(1 - \ln b) - (a-b) \ln(1-\omega)) \ln \omega - (a+b) (\ln(a b) - 2) \ln(1+\omega) \right), \\ 0 < b < a, \quad\omega = \frac b a.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.