Integral $intlimits^{infty}_0frac{tan^{-1}t }{(1+t)^{n+1}} dt$

Question

I'm having a good amount of trouble evaluating this: $$intlimits^{infty}_0frac{tan^{-1}(t)dt}{(1+t)^{n+1}}, n>0$$
Here are some methods I've tried:
$$intlimits^{infty}_0frac{tan^{-1}(t)dt}{(1+t)^{n+1}}=frac1nintlimits^{infty}_0frac{dt}{(1+t^2)(1+t)^{n}}$$
using integration by parts. I then tried more integration by parts, residue theorem, and expanding into a power series but failed. I did however use partial fractions for $n=2$ to get $1/4$.
$$intlimits^{infty}_0frac{tan^{-1}(t)dt}{(1+t)^{n+1}}=frac{pi}{2n}-intlimits^{infty}_0frac1{(1+t)^{n+1}}intlimits^{infty}_0frac{sin(x)}xe^{-xt}dxdt=frac{pi}{2n}-intlimits^{infty}_0frac{sin(x)}xE_{n+1}(x)e^{-x}dx$$
using the Laplace Transform of $text{sinc}(x)$ and the $E_n$-function.
$$intlimits^{infty}_0frac{tan^{-1}(t)dt}{(1+t)^{n+1}}=intlimits^1_0frac{tan^{-1}(t)dt}{(1+t)^{n+1}}+intlimits^1_0frac{cot^{-1}(t)t^{n-1}dt}{(1+t)^{n+1}}=intlimits^1_0frac{tan^{-1}(t)left(1-t^{n-1}right)dt}{(1+t)^{n+1}}+frac{pi}{2^{n+1}n}$$
This one I felt the best about and its also where I got that at $n=1$ the integral is $pi/4$, but I was unable to go further.
Update: I had a couple more attempts, one of which I posted as an answer, after Claude Leibovici's idea reminded me you can do partial fractions on $frac1{(1+x^2)(1+x)^n}$.
Notice that if we write
$$frac1{(1+x^2)(1+x)^n}=frac{1+x}{1+x^2}-frac{a_0+a_1x+dots+a_{m-1}x^{m-1}}{(1+x)^n}$$
then the coefficients $a_k$ follow the pattern
$$a_0=0, a_1=C_1^{n+1}, a_2=C_2^{n+1}-a_0, a_3=C_3^{n+1}-a_1, a_4=C_4^{n+1}-a_2dots$$
The only problem is that this sequence is always infinite and the power series does not converge on all of $[0,infty)$, so I believe the coefficients of $1$ and $x$ in the numerator of $frac{1+x}{1+x^2}$ could be changed to avoid this.

Quanto · Accepted Answer

Note that $intlimits^{infty}_0frac{tan^{-1}t}{(1+t)^{n+1}}=frac1nI_n$, where
$$I_n=intlimits^{infty}_0frac{dt}{(1+t^2)(1+t)^{n}}$$
The integrand can be decomposed iteratively as
$$A_n(t)= frac{A_{n-1}}{1+t}=frac{1}{(1+t^2)(1+t)^{n}}
=frac{a_n-b_n t}{1+t^2}+ sum_{k=1}^{n}frac{b_{n-k+1}}{(1+t)^k}tag1
$$
where the coefficients satisfy the iterative relationships
$$a_n=frac{a_{n-1}-b_{n-1}}2,>>>>> b_n=frac{a_{n-1}+b_{n-1}}2tag2$$
Recognize $a_0=1$,  $b_0=0$
and compare
$$cos frac{npi}4= frac1{2^{frac12}}left(cos frac{(n-1)pi}4-sinfrac{(n-1)pi}4right)
$$
$$sin frac{npi}4= frac1{2^{frac12}}left(cos frac{(n-1)pi}4+sinfrac{(n-1)pi}4right)
$$
with (2) to get
$$a_n=frac1{2^{frac n2} }cosfrac{npi}4,>>>>>
 b_n=frac1{2^{frac n2} }sinfrac{npi}4tag3
$$
Then, integrate $A_n(t)$ in (1) to obtain
$$I_n= int_0^infty A_n(t)dt =frac{pi a_n}2+sum_{j=1}^{n-1}frac{b_{j}}{n-j}
$$
Substitute the coefficients (3) to arrive at the result
$$I_n = fracpi{2^{frac{n+1}2}}cosfrac{npi}4
+ sum_{j=1}^{n-1}frac{1}{(n-j) 2^{frac j2}}sinfrac{jpi}4
$$
Listed below are the first few integral values
begin{align}
& I_1 =fracpi4 \ & I_2 =frac12\
 & I_3 =frac34-fracpi8\
 & I_4 =frac23-fracpi8\
 & I_5 =frac{5}{12}-fracpi{16}\
end{align}

Chunky Norris · Answer

I was able to get some kind of recurrence relation for
$$I_n=intlimits_0^{infty}frac{dt}{(1+t^2)(1+t)^n},$$
but I am not satisfied with it since you can't really do anything with it. It's still kind of an answer but I'll accept a better one.
First substitute $tmapstofrac1t$ so that
$$I_n=intlimits_0^{infty}frac{t^ndt}{(1+t^2)(1+t)^n}.$$
Notice that you can modify the binomal expansion so it is similar to partial fractions:
$$begin{align*}
(1+t)^n&=sum_{k=0}^nleft(begin{matrix}n\kend{matrix}right)t^k\
left(1+frac{-1}tright)^n&=sum_{k=0}^nleft(begin{matrix}n\kend{matrix}right)left(frac{-1}tright)^k\
frac{t^n}{(1+t)^n}&=sum_{k=0}^nleft(begin{matrix}n\kend{matrix}right)left(frac{-1}{1+t}right)^k.\
end{align*}$$
It then follows that
$$I_n=intlimits_0^{infty}frac1{(1+t^2)}sum_{k=0}^nleft(begin{matrix}n\kend{matrix}right)left(frac{-1}{1+t}right)^kdt
=sum_{k=0}^nleft(begin{matrix}n\kend{matrix}right)(-1)^kI_k.$$
$$impliesboxed{(1-(-1)^n)I_n=sum_{k=0}^{n-1}left(begin{matrix}n\kend{matrix}right)(-1)^kI_k}$$
Unfortunately it sucks and the most I was able to do with it was find $I_3=3/4-pi/8$ from already knowing $I_0=pi/2,$ $I_1=pi/4,$ and $I_2=1/2$.
There may be some merit in the study of
$$I_n=frac12intlimits_0^{infty}frac{(1+t^n)dt}{(1+t^2)(1+t)^n}$$
or maybe splitting the interval into $[0,1]$ and $[1,infty)$.

Claude Leibovici · Answer

This is not an answer but it is too long for comments.
For the computation of
$$I_n=intlimits^{infty}_0frac{dt}{(1+t^2)(1+t)^{n+1}}$$ it is amazing that a CAS gives a solution in terms of a generalized hypergeometric function which works very fine ... except when $n$ is an integer !
What I think is that writing
$$(1+t^2)(1+t)^{n+1}=(t+i)(t-i)(1+t)^{n+1}$$ and using partial fraction could be a solution. For example, for $n=3$, the integrand is
$$-frac{1+i}{8(t+i)}-frac{1-i}{8(t-i)}+frac{1}{4 (t+1)}+frac{1}{2
   (t+1)^2}+frac{1}{2 (t+1)^3}$$ and
$$int Big[frac{1+i}{8(t+i)}+frac{1-i}{8(t-i)}Big],dt=frac{1}{8} log left(t^2+1right)+frac{1}{4} tan ^{-1}(t)$$ For $n=4$ , the integrand is
$$frac{i}{8 (t-i)}-frac{i}{8 (t+i)}+frac{1}{4 (t+1)^2}+frac{1}{2 (t+1)^3}+frac{1}{2   (t+1)^4}$$
$$int Big[frac{i}{8 (t-i)}-frac{i}{8 (t+i)}Big],dt=-frac{1}{4} tan ^{-1}(t)$$ and, obviously,  the coefficients of the terms $frac{1}{ tpm i}$ are complex numbers if $n$ is odd and pure imaginary numbers if $n$ is even.
Probably, the two cases could be separatly studied.
All these integrals are in the form $I_n=a_n+b_npi$ but the $b_n$'s are all zero for $n=4k+2$

Integral $intlimits^{infty}_0frac{tan^{-1}t }{(1+t)^{n+1}} dt$

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