Vtyjkyuk

Motivated by the easier integral
$$ int limits_0^infty left[frac1x^2 - frac1x sinh(x)right] mathrmd x = ln(2) , ,$$
I have been trying to compute

$$ I equiv int limits_0^pi/2 left[frac1x sin(x) - frac1x^2 right] mathrmd x approx 0.29172334953491321 , .$$

I have not found a closed-form expression yet and inverse symbolic calculators do not give any results either. However, a few other representations for $I$ can be derived using the following methods:

1. 
   

   Laurent series

   
   
   

   The Laurent series of the cosecant function is given by $$csc(x) = frac1x + sum limits_k=1^infty fraclvert mathrmB_2krvert (4^k - 2)(2k)! x^2k-1$$ in terms of the Bernoulli numbers $(mathrmB_n)_n in mathbbN_0$ and has radius of convergence $pi$ , so we can integrate termwise to obtain $$ tag1 I = sum limits_k=1^infty fraclvert mathrmB_2krvert left[2-4^-(k-1)right] pi^2k-1(2k-1)(2k)! , .$$

   
   


2. 
   

   Pole expansion

   
   
   

   The series $$ csc(x) = frac1x + 2 x sum limits_n=1^infty frac(-1)^n-1pi^2 n^2 - x^2$$ yields $$tag2 I = frac1pi sum limits_n=1^infty frac(-1)^n-1n lnleft(frac2n+12n-1right) , .$$ Expanding the logarithm only leads to $$ tag3 I = frac1pi sum limits_k=1^infty fraceta(2k)(2k-1) 4^k-1 , ,$$ which reduces to $(1)$ when the special values of the eta functions are used. Summation by parts turns $(2)$ into
   $$ tag4 I = frac4pi sum limits_n=1^infty frac(-1)^n-1 (2n+1) ln(2n+1)(2n+1)^2 -1 , . $$
   This can also be written as
   $$ tag5 I = frac4pi beta'(1) + frac1pi sum limits_n=1^infty frac(-1)^n-1 ln(2n+1)2n^3+3n^2+n , ,$$
   where $beta$ is the Dirichlet beta function (there is a reasonably nice expression for $beta'(1)$).

   
   


3. 
   

   Integration by parts

   
   
   

   There are several possible ways to integrate by parts. One of them shows that $$ tag6 I = frac2pi ln left(frac4piright) + frac12 int limits_0^pi/4 fracln[tan(t)/t]t^2 , mathrmd t $$ holds. I am not sure how to proceed from here though. Plugging in the Maclaurin series of $ln[tan(t)/t]$ reproduces $(1)$ .

   
   


4. 
   

   Contour integration (due to eyeballfrog)

   
   
   

   As demonstrated in eyeballfrog's answer we also have $$ tag7 I = frac2pi - int limits_0^infty fract1+t^2 , operatornamesechleft(fracpi2 tright) , mathrmd t , .$$ Using the pole expansion of $operatornamesech$ yields $(4)$ again.

   
   

That is all I have at the moment, so my question is:

Is it possible to find a closed-form expression for the value of $I$ or can we at least rewrite any of the integral or series representation in terms of a suitable special function?

Well here's a start. Consider the contour around the rectangle with corners at $0, pi/2, pi/2+iR,iR$. $1/[xsin(x)]-1/x^2$ has no poles in this region, so the integral around the contour must be 0. Parameterizing the integrals along each side of the contour then gives
$$
int_0^pi/2left[frac1xsin x-frac1x^2right]dx + iint_0^Rleft[frac1(pi/2+iy)sin (pi/2+iy)-frac1(pi/2+iy)^2right]dy - int_0^pi/2left[frac1(x+iR)sin(x+iR)-frac1(x+iR)^2right]dx -i int_0^Rleft[frac1iysin(iy)-frac1(iy)^2right]dy = 0
$$
The first integral is the one we want. The third integral vanishes as $Rrightarrowinfty$, while the fourth integral is just $int_0^infty[1/y^2 - 1/(ysinh y)]dy = ln 2$ in that limit. So we have in the limit that $Rrightarrowinfty$,
$$
int_0^pi/2left[frac1xsin x-frac1x^2right]dx + ileft(int_0^inftyleft[frac1(pi/2+iy)sin (pi/2+iy)-frac1(pi/2+iy)^2right]dy - ln 2right) = 0
$$
Since the first integral is purely real, the $ln 2$ term must cancel out the real part of the second integral, and we have
$$
int_0^pi/2left[frac1xsin x-frac1x^2right]dx =int_0^inftymathrmImleft[frac1(pi/2+iy)sin (pi/2+iy)-frac1(pi/2+iy)^2right]dy
$$
Actually expanding the imaginary part and doing various algebra things gives
$$
int_0^pi/2left[frac1xsin x-frac1x^2right]dx = frac2pi - int_0^infty fract1+t^2mathrmsechleft(fracpi2tright)dt
$$
Despite the relative simplicity of this last integral, Mathematica won't do it and I can't find it or things that might lead to it in Gradshteyn and Rhyzik. Anyone have an idea of where to go from here?

Without the $t$ in the numerator, there is the closed-form evaluation for the eyeballfrog-like integral
$$ int_0^infty fracdta^2+t^2,textsech(pi t/2) = frac12aBig(psi(fraca+34) - psi(fraca+14) Big) .$$
I suspect that 'missing t' will prevent the answer from being solved in closed form. There are variants with $t/sinh(pi t/2)$ and sech-squared, but always the integrand is even. Nevertheless, with some calisthenics one may derive
$$ int_0^infty dt fract1+t^2textsech(pi t/2) = 2int_0^infty
cos(2pi u),big(psi(3/4+u) - psi(1/4+u) big) du .$$
This provides a little hope because it is known that
$$int_0^infty
cos(2pi u ,x),big(psi(1+u) - log(u) big) du = frac12big(psi(1+x) - log(x) big).$$
If only the argument of the digamma function could be replaced with $1+u to a+u$ and the new right-hand side possessed a closed form evaluation, then the problem would be solved.