Integral Integral 4

We begin by factorising

$\displaystyle 1+\frac { 1 }{ x } +x+\frac { 1 }{ { x }^{ 2 } } =\left( 1+x \right) \left( 1+\frac { 1 }{ { x }^{ 2 } } \right)$

$\displaystyle \int _{ 0 }^{ 1 }{ \ln { ^{ 2 } } { (1+x+\frac { 1 }{ x } +\frac { 1 }{ { x }^{ 2 } } ) }dx } =\int _{ 0 }^{ 1 }{ \left( \ln^{2}(1+x)+2\ln(1+x)\ln(1+\frac{1}{x^{2}})+\ln^{2}(1+\frac{1}{x^{2}}) \right) dx }$

Now, we integrate each term separately

$\displaystyle \int _{ 0 }^{ 1 }{ \left( \ln { ^{ 2 } } (1+x) \right) dx } =x\ln^{2}(1+x)-2\int^{1}_{0}{\dfrac{2\ln(1+x)}{1+x} dx }=\ln^{2}{2}-2 \left[ \int^{1}_{0}{(\ln{1+x}-\dfrac{\ln{1+x}}{1+x})} \right ]=2(\ln{2}-1)^{2}$

$\displaystyle \int _{ 0 }^{ 1 }{ \left( \ln { ^{ 2 } } (1+\dfrac{ 1 }{ x^{ 2 } }) \right) dx } =x\ln { ^{ 2 } } (1+\frac{ 1 }{ x^{ 2 } })+4\int _{ 0 }^{ 1 }{ \frac { \ln{1+x^{2}}-2\ln{x} }{ 1+x^{2} } dx }$

Now

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { \ln { 1+x^{ 2 } } -2\ln { x } }{ 1+x^{ 2 } } dx } =\int _{ 1 }^{ \infty }{ \frac { \ln { 1+x^{ 2 } } }{ 1+x^{ 2 } } dx }$

So,

$\displaystyle \int _{ 0 }^{ \infty }{ \frac { \ln { 1+x^{ 2 } } }{ 1+x^{ 2 } } dx } +\int _{ 0 }^{ 1 }{ \frac { -2\ln { x } }{ 1+x^{ 2 } } dx } =2\int _{ 1 }^{ \infty }{ \frac { \ln { 1+x^{ 2 } } }{ 1+x^{ 2 } } dx }$

To evaluate each term, we first let

$\displaystyle J(a)=\int _{ 0 }^{ \infty }{ \frac { \ln { 1+{ a }^{ 2 }x^{ 2 } } }{ 1+x^{ 2 } } dx }$

$\displaystyle J'(a)=\int _{ 0 }^{ \infty }{ \frac { { 2ax }^{ 2 } }{ (1+x^{ 2 })(1+{ a }^{ 2 }{ x }^{ 2 }) } dx } =\frac { 2a }{ 1{ -a }^{ 2 } } \left( \int _{ 0 }^{ \infty }{ \frac { 1 }{ (1+{ a }^{ 2 }{ x }^{ 2 }) } dx } -\int _{ 0 }^{ \infty }{ \frac { 1 }{ (1+x^{ 2 }) } dx } \right) =\frac { \pi }{ 1+a }$

So,

$\displaystyle J(1)=\pi\ln{2}$

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { \ln { x } }{ 1+{ x }^{ 2 } } dx } =\ln{x}\tan^{-1}{x}-\int _{ 0 }^{ 1 }{ \frac {\tan^{-1}{x}}{ x } dx } =-G$ (The last can be done easily by expanding the infinite series for $\tan^{-1}{x}$ )

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { \ln { 1+x^{ 2 } } -2\ln { x } }{ 1+x^{ 2 } } dx }=\dfrac{\pi\ln{2}}{2}+G$

Now, the last one ;D

$\displaystyle \int _{ 0 }^{ 1 }{ \ln { x } \ln { 1+\frac { 1 }{ { x }^{ 2 } } } dx } =x\ln { x } \ln { 1+\frac { 1 }{ { x }^{ 2 } } } -\int _{ 0 }^{ 1 }{ \frac { x }{ 1+x } \ln { 1+\frac { 1 }{ { x }^{ 2 } } } dx } +2\int _{ 0 }^{ 1 }{ \frac { \ln { 1+x } }{ 1+{ x }^{ 2 } } dx }$

By letting

$\displaystyle K(a)=\int _{ 0 }^{ 1 }{ \frac { \ln { 1+ax } }{ 1+{ x }^{ 2 } } dx }$

We have

$\displaystyle K'(a)=\int _{ 0 }^{ 1 }{ \frac { x }{ (1+{ x }^{ 2 })(1+ax) } dx } =\frac { a }{ 1+{ a }^{ 2 } } \int _{ 0 }^{ 1 }{ \frac { 1+\frac { x }{ a } }{ (1+{ x }^{ 2 }) } -\frac { 1 }{ (1+ax) } dx } =\frac { a }{ 1+{ a }^{ 2 } } \left( \frac { \pi }{ 4 } +\frac { \ln { 2 } }{ 2a } -\frac { \ln { 1+a } }{ a } \right)$

So,

$\displaystyle K(1)-K(0)=\dfrac{\pi\ln{2}}{4}-K(1)$ Note that $K(0)=0$ Then,

$\displaystyle \int _{ 0 }^{ 1 }{ \frac { \ln { 1+x } }{ 1+{ x }^{ 2 } } dx } =\frac { \pi \ln { 2 } }{ 8 }$

$\displaystyle \int _{ 0 }^{ 1 }{ \ln { x } \ln { 1+\frac { 1 }{ { x }^{ 2 } } } dx } =\ln^{2}{2}+\dfrac{\pi\ln{2}}{4}-\int_{0}^{1}{(1-\dfrac{1}{1+x})\ln{(1+x^{2})}dx}+2 \int_{0}^{1}{(1-\dfrac{1}{1+x})\ln{x}dx}$

Now we just need to integrate the last 4 terms,( from here on I deliberately leave out the limit of integration for the sake of simplicity , therefore the limit is well known to be from 0 to 1)

$\displaystyle \int\ln{1+x^{2}}=\ln{2}-2+\dfrac{\pi}{2}dx$ (The integral here is easy, so, I leave it as an exercise.)

$\displaystyle \int\ln{x}dx=-1$

$\displaystyle \int{\dfrac{\ln{x}}{1+x}}=-\int{\dfrac{\ln{1+x}}{x}}=-\dfrac{\pi^{2}}{12}$ (This integral is done by expanding the series for logarithm and knowing the fact that $\displaystyle \sum_{1}^{\infty}{\dfrac{1}{n^{2}}}=\dfrac{\pi^{2}}{6}$ )

$\displaystyle \int{\dfrac{\ln{1+x^{2}}}{1+x}dx}=-\dfrac{\pi^{2}}{48}+\dfrac{3\ln^{2}{2}}{4}$

The last integral can be done by lettting

$\displaystyle L(a)=\int{\dfrac{\ln{1+a^{2}x^{2}}}{1+x}dx}$

$\displaystyle L'(a)=\dfrac{2a}{1+a^{2}}(\int{\dfrac{1}{1+x}+\dfrac{x-1}{1+a^{2}x^{2}}dx})$ (Recall $\displaystyle L(0)=0$ )

We get,

$\displaystyle L(1)=\ln^{2}{2}-\dfrac{\pi^{2}}{16}+\int{\dfrac{\ln(1+x^{2})}{x(1+^{2})}dx}=\ln^{2}{2}-\dfrac{\pi^{2}}{16}+\int{\dfrac{\ln(1+x^{2})}{x}dx}-\int{\dfrac{x\ln(1+x^{2})}{1+x^{2}}dx}$

$\displaystyle L(1)=\ln^{2}{2}-\dfrac{\pi^{2}}{16}-\dfrac{\ln^{2}{2}}{4}+\dfrac{\pi^{2}}{24}$

Combining all these togather

$\displaystyle \int _{ 0 }^{ 1 }{ \ln { ^{ 2 } } { (1+x+\frac { 1 }{ x } +\frac { 1 }{ { x }^{ 2 } } ) }dx }=2+4G+\dfrac{7\pi^{2}}{24}-6\ln{2}+\dfrac{13\ln^{2}}{2}+\pi(\dfrac{5\ln{2}}{2}-1)$

So, $\displaystyle a+b+c+d+f+g+h+j+k+m=\boxed{66}$