TITLE: Calculate in closed form $\int_{0}^{\infty}({1-12y^2)(1+4y^2)^{-t}} ~dy,$ for a fixed positive number $t.$
QUESTION [0 upvotes]: Question: What is the closed form of 
  $$\int_{0}^{\infty}({1-12y^2)(1+4y^2)^{-t}} ~dy
$$ for a fixed positive number $t$?


For short handedness I write $F_t(y)=(1-12y^6)(1+4y^2)^{-t}.$ The table below shows computed values of the integral for small values of $t:$
$$
\begin{array}{c|lcr}
t & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9  \\
\hline
\int_0^\infty F_t(y)~dy & \frac{\pi}{4} & 0 & \frac{\pi}{32} & \frac{5\pi}{128} & \frac{21\pi}{512} & \frac{21\pi}{512} & \frac{168\pi}{4096} & \frac{1287\pi}{32768} \\
\end{array}
$$

Here is an attempted solution for the value $t=2.$ This was proven with help from Wolfram Alpha and Wikipedia. I suspect a more clear and concise solution exist. 
By partial fraction expansion I have that:
\begin{align}
\int_0^\infty F_t(y)~dy&=\int_{0}^{\infty}(1-12y^2)(1+4y^2)^{-t} ~dy\\
&=-3\int_0^\infty (4y^2+1)^{-2}~dy+4\int_0^\infty (4y^2+1)~dy\\
\end{align}
Working on the first integral of the RHS of the equality above I substitute $u=2y$ and so $~du=~2dt$ and apply the fundamental theorem of calculus-  noting that the antiderivative of $(u^2-1)^{-1}=\tan^{-1}(u).$ In particular I now have
$$
\lim_{b\to \infty}\left(-\frac{3}{2}\tan^{-1}(u)\right)\Bigg|_{0}^b+4\int_0^\infty (4y^2+1)^{-2}~dy;
$$
which is equal to 
$$
-\frac{3}{4}\pi+4\int_0^\infty (4y^2+1)^{-2}~dy;
$$
Now working on the integral above I substitute $y=\frac{1}{2}\tan(u)$ and so $du=\frac{1}{2}\sec^2(u)~du$ in which case $(4y+1)^2=(\tan^2(u)+1)^2=\sec^4(u).$ This substitution is invertible whenever $0<u<\frac{\pi}{2}$ with inverse $u=\tan^{-1}(2y).$ This gives a new lower and upper bound for $u$: $0$ and $\frac{\pi}{2}$ respectively. I now have 
$$
-\frac{3}{4}\pi+2\int_0^{\frac{\pi}{2}}\cos^2(u)~du;
$$
Rewriting $\cos^2(u)$ as $\frac{1}{2}(2u)+\frac{1}{2}$ we have
$$
-\frac{3}{4}\pi+\frac{1}{2}\int_0^{\frac{\pi}{2}}\cos(2u)~du+\int_0^{\frac{\pi}{2}}1~du
$$
A final substitution, $v=2u$ so that $dv=2du,$ yields the lower bound and upper bounds for $v$ equal to $0$ and $\pi$ respectively. Applying the fundamental theorem of calculus again this time noting the antiderivative of $\cos(u)=\sin(u),$ yields
$$
-\frac{3}{4}\pi+\frac{1}{2}\sin(v)\Bigg|_0^{\pi}+\int_0^{\frac{\pi}{2}}1~du
$$
Observe: $\frac{1}{2}\sin(v)\Big|_0^{\pi}=\frac{1}{2}(\sin(\pi)-\sin(0)=0$ and 
$\int_0^{\frac{\pi}{2}}1~du$ is equal to $\frac{\pi}{2}.$ Then
$$
-\frac{3}{4}\pi+\frac{1}{2}\pi=-\frac{\pi}{4}.
$$

It appears that $\int_0^\infty F_t(y)~dy=q_t\frac{\pi}{4}$ for some, possibly negative, irrational number $q_t;$ which is dependent on $t.$ The values of $q_t$ are really what I am after.

REPLY [3 votes]: Use $y=2 \tan u, dy=2 \mbox{sec}^2 u du$ and then Beta-integral in terms of Gamma functions, we get $$I=\frac{(t-3) \sqrt{\pi}~ \Gamma(t-3/2)}{4 \Gamma(t)}, t>3/2.$$
{"set_name": "stack_exchange", "score": 0, "question_id": 3285097}