Vrftsjtryk

I am in the process of proving
$$I=int_0^infty frac{arctan x}{x^4+x^2+1}mathrm{d}x=frac{pi^2}{8sqrt{3}}-frac23G+fracpi{12}log(2+sqrt{3})$$
And I have gotten as far as showing that
$$2I=frac{pi^2}{4sqrt{3}}+J$$
Where
$$J=int_0^infty logbigg(frac{x^2-x+1}{x^2+x+1}bigg)frac{mathrm{d}x}{1+x^2}$$
Then we preform $x=tan u$ to see that
$$J=int_0^{pi/2}logbigg(frac{2+sin2x}{2-sin2x}bigg)mathrm dx$$
Which I have been stuck on for the past while. I tried defining
$$k(a)=int_0^{pi/2}log(2+sin2ax)mathrm dx$$
Which gives
$$J=k(1)-k(-1)$$
Then differentiating under the integral:
$$k'(a)=2int_0^{pi/2}frac{xcos2ax}{2+sin2ax}mathrm dx$$
We may integrate by parts with $u=x$ to get a differential equation
$$ak'(a)+k(a)=fracpi2log(2+sinpi a)$$
With initial condition
$$k(0)=fracpi2log2$$
And from here I have no idea what to do.

I also tried tangent half angle substitution, but that just gave me the original expression for $J$.

I'm hoping that there is some really easy method that just never occurred to me... Any tips?

Edit

As was pointed out in the comments, I could consider
$$P(a)=frac12int_0^pi log(a+sin x)mathrm dx\Rightarrow P(0)=-fracpi2log2$$
And
$$
begin{align}
Q(a)=&frac12int_0^pi log(a-sin x)mathrm dx\
=&frac12int_0^pilog[-(-a+sin x)]mathrm dx\
=&frac12int_0^pibigg(log(-1)+log(-a+sin x)bigg)mathrm dx\
=&frac{ipi}2int_0^pimathrm{d}x+frac12int_0^pilog(-a+sin x)mathrm dx\
=&frac{ipi^2}2+P(-a)
end{align}
$$
Hence
$$J=P(2)-Q(2)=P(2)-P(-2)-frac{ipi^2}2$$
So now we care about $P(a)$. Differentiating under the integral, we have
$$P'(a)=frac12int_0^pi frac{mathrm{d}x}{a+sin x}$$
With a healthy dose of Tangent half angle substitution,
$$P'(a)=int_0^infty frac{mathrm{d}x}{ax^2+2x+a}$$
completing the square, we have
$$P'(a)=int_0^infty frac{mathrm{d}x}{a(x+frac1a)^2+g}$$
Where $g=a-frac1a$. With the right trigonometric substitution,
$$P'(a)=frac1{sqrt{a^2+1}}int_{x_1}^{pi/2}mathrm{d}x$$
Where $x_1=arctanfrac1{sqrt{a^2+1}}$. Then using
$$arctanfrac1x=fracpi2-arctan x$$
We have that
$$P'(a)=frac1{sqrt{a^2+1}}arctansqrt{a^2+1}$$
So we end up with something I don't know how to to deal with (what a surprise)
$$P(a)=intarctansqrt{a^2+1}frac{mathrm{d}a}{sqrt{a^2+1}}$$
Could you help me out with this last one? Thanks.

$$J=int_0^{pi/2}lnleft(frac{2+sin2x}{2-sin2x}right)mathrm dxoverset{2x=t}=frac12 int_0^pi lnleft(frac{1+frac12sin t}{1-frac12sin t}right)mathrm dt=int_0^frac{pi}{2}lnleft(frac{1+frac12sin x}{1-frac12sin x }right)mathrm dx$$
$${lnleft(frac{1+ysin x}{1-ysin x}right)+c=2intfrac{sin x}{1-y^2sin^2x}dy}Rightarrowlnleft(frac{1+frac12sin t}{1-frac12sin t }right)=2int_0^frac{1}{2}frac{sin x}{1-y^2sin^2x}dy tag1$$

By Fubini's theorem we may interchange the order of the integrals and using $(1)$ we get: $$J=2int_0^frac{pi}{2}int_0^frac12 frac{sin x}{1-y^2sin^2x}dydx=2int_0^frac12int_0^frac{pi}{2} frac{sin x}{1-y^2sin^2x}dxdy=$$
$$=2int_0^frac12int_0^frac{pi}{2} frac{sin x}{y^2cos^2x +1-y^2} dxdy=2int_0^frac12frac{1}{ysqrt{1-y^2}}arctanleft(frac{yx}{sqrt{1-y^2}}right)bigg|_0^1 dy=$$
$$=2int_0^frac12frac{1}{ysqrt{1-y^2}}arctanleft(frac{y}{sqrt{1-y^2}}right)dyoverset{y=sin x}=2int_0^frac{pi}{6}frac{x}{sin x}dx=$$
$$=2int_0^{frac{pi}{6}} x left(lnleft(tan frac{x}{2}right)right)'dx=2x lnleft(tan frac{x}{2}right)bigg|_0^{frac{pi}{6}} -2{int_0^{frac{pi}{6}} lnleft(tan frac{x}{2}right)dx}=$$
$$overset{frac{x}{2}=t}=frac{pi}{3}ln(2-sqrt 3) -4int_0^frac{pi}{12}ln (tan t)dt=-frac{pi}{3}ln(2+sqrt 3) +frac{8}{3}G$$ $G$ is the Catalan's constant and for the last integral see here.

Also note that you should have: $$2I=frac{pi^2}{4sqrt 3}- int_0^inftyfrac{(x^2-1)arctan x}{x^4+x^2+1}dx=frac{pi^2}{4sqrt 3}-frac12underset{=J}{int_0^infty lnbigg(frac{x^2-x+1}{x^2+x+1}bigg)frac{dx}{1+x^2}}$$

As an alternative employ Feynman's trick for $,displaystyle{I(a)=int_0^frac{pi}{2}lnleft(frac{1+asin x}{1-asin x}right)dt},$ to get:
$$I'(a)=int_0^frac{pi}{2} left(frac{sin x}{1+asin x}+frac{sin x}{1-asin x}right)dx =2int_0^frac{pi}{2} frac{sin x}{1-a^2sin^2 x}dx=$$
$$=2int_0^frac{pi}{2} frac{sin x}{a^2cos^2 x + left(sqrt{1-a^2}right)^2}dx=frac{2}{a^2}int_0^1 frac{d(cos x)}{cos^2x +left(frac{sqrt{1-a^2}}{a}right)^2}=$$
$$=frac{2a}{a^2sqrt{1-a^2}}arctanleft(frac{ax}{sqrt{1-a^2}}right)bigg|_0^1=frac{2}{asqrt{1-a^2}}arctanleft(frac{a}{sqrt{1-a^2}}right)$$
$$I(0)=0 Rightarrow J=I(tfrac12)=int_0^frac12 frac{2}{asqrt{1-a^2}}arctanleft(frac{a}{sqrt{1-a^2}}right)overset{a=sin x}=2int_0^frac{pi}{6}frac{x}{sin x}dx$$
And the rest part is found above.

A Possible way: Consider
$$I(a)=int_{0}^{+infty}frac{arctan(ax)}{1+x^2+x^4} dx$$
and
$$I'(a)=int_{0}^{+infty}frac{x}{(1+x^2+x^4)(1+x^2a^2)}dx=int_{0}^{+infty}frac{1}{(1+y+y^2)(1+a^2y)}dx$$
and

$$frac{1}{(1+y+y^2)(1+a^2y)}= frac{-a^2y-a^2+1}{(a^4-a^2+1)(1+y+y^2)}+frac{a^4}{(a^4-a^2+1)(ay^2+1)}$$

we can also calculate $I'(a)$ by complex integration (if you've learned that).

Thanks to Dylan for his advice.

Result

I find that the integral has a closed form given by

$$i = intlimits_0^{pi/2}logbigg(frac{2+sin2x}{2-sin2x}bigg)mathrm dx = frac{1}{3} left(8 C-pi log left(2+sqrt{3}right)right) simeq 1.06346tag{1}$$

where

$$C = sum _{k=1}^{infty } frac{(-1)^{k+1}}{(2 k-1)^2} simeq 0.915966$$

is Catalan's constant.

Heuristic derivation

Notice trivially that because of the symmtery of the integrand the integral can be written as twice the integral from $0$ to $frac{pi}{4}$ which we shall utilize in what follows.

The basic idea is the series expansion

$$log left(frac{1+z}{1-z}right)=2tanh ^{-1}(z) = 2 sum _{k=1}^{infty } frac{z^{2 k-1}}{2 k-1},|z|<1 tag{2}$$

The integral is then to be done over the odd powers of the $sin$ with the result

$$int_0^{frac{pi }{4}} sin ^{2 k-1}(2 x) , dx = frac{sqrt{pi } Gamma (k)}{4 Gamma left(k+frac{1}{2}right)}tag{3}$$

Assembling the pieces the sum to be taken to represent $i$ becomes

$$i_s = sum _{k=1}^{infty } frac{sqrt{pi } Gamma (k)}{(2 k-1) 2^{2 k-1} Gamma left(k+frac{1}{2}right)}tag{4}$$

and this sum is immediately computed by Mathematica to give the compact result $(1)$.

Let us make the sum more transparent using the chain

$$frac{sqrt{pi } Gamma (k)}{Gamma left(k+frac{1}{2}right)}=Bleft(frac{1}{2},kright)=int_0^1 frac{t^{k-1}}{sqrt{1-t}} , dttag{5}$$

and doing the sum under the integral

$$sum _{k=1}^{infty } frac{t^{k-1}}{(2 k-1) 2^{2 k-1}}=frac{tanh ^{-1}left(frac{sqrt{t}}{2}right)}{sqrt{t}}tag{6}$$

leads finally to the integral

$$int_0^1 frac{tanh ^{-1}left(frac{sqrt{t}}{2}right)}{sqrt{t} sqrt{1-t}} , dttag{7}$$

for which Mathematica again quickly gives (1).

But there must be a shorter way ... yes, it is, substituting $sin (2 x)=sqrt{t}$ in the original integral gives (7) directly.