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Let $0<b<a$ and define $$J(a,b)=\int_0^1\frac{\arctan\sqrt{t^2+a}}{(t^2+b)\sqrt{t^2+a}}dt.\tag1$$

I am seeking a closed form for $J(a,b)$.


I was motivated to find a closed form for $(1)$ after seeing the approach taken in this answer and thinking that it could be generalized.

I have actually succeeded in finding a closed form for the special case $(a,b)=(a,\tfrac{a}{2})$:

$$J(a,\tfrac{a}{2})=\frac{\pi^2}{2a}-\frac{\pi}{a}\arctan\sqrt{a+1}-\frac1a\arctan^2\sqrt{\frac{2}{a}},$$ which comes from the more general result $$J(a,b)+J(a,a-b)=\phi_2(a,b)-\phi_1(b)\phi_1(a-b)+\phi_2(a,a-b),\tag2$$ where $$\phi_1(x)=\int_0^1\frac{dt}{x+t^2}=\int_1^\infty\frac{dt}{1+xt^2}=\frac1{\sqrt x}\arctan\frac1{\sqrt x},$$ and $$\phi_2(a,b)=\frac{\pi}{2\sqrt{b(a-b)}}\arctan\sqrt{\frac{a-b}{b(a+1)}}.$$ I will supply a proof below.


Proof of $(2)$. Let $$f(z)=\int_0^1\frac{\arctan(z\sqrt{t^2+a})}{(t^2+b)\sqrt{t^2+a}}dt,$$ so that $J(a,b)=f(1)$. Then clearly $$J(a,b)=f(1)=\lim_{z\to\infty}f(z)-\int_1^\infty f'(z)dz.$$ Then since $\lim_{z\to\infty}\arctan(xz)=\pi/2$ for $x>0$, we have $$J(a,b)=\frac\pi2\int_0^1\frac{dt}{(t^2+b)\sqrt{t^2+a}}-\int_1^\infty\int_0^1\frac{dt}{(t^2+b)(z^2t^2+az^2+1)}dz.$$ The first integral is relatively simple: $$\begin{align} \int_0^1\frac{dt}{(t^2+b)\sqrt{t^2+a}}&=\int_0^{1/\sqrt{a}}\frac{du}{(au^2+b)\sqrt{u^2+1}}\qquad[t\mapsto u\sqrt{a}]\\ &=\int_0^{\arctan 1/\sqrt{a}}\frac{\sec^2x\,dx}{(a\tan^2x+b)\sqrt{\tan^2x+1}}\qquad[u\mapsto \tan x]\\ &=\int_0^{\arctan 1/\sqrt{a}}\frac{\cos x\,dx}{b+(a-b)\sin^2x}\\ &=\int_0^{1/\sqrt{a+1}}\frac{dt}{b+(a-b)t^2}\qquad [\sin x\mapsto t]\\ &=\left.\frac{1}{\sqrt{b(a-b)}}\arctan\left(t\sqrt{\frac{a-b}{b}}\right)\right|_0^{1/\sqrt{a+1}}\\ &=\frac{1}{\sqrt{b(a-b)}}\arctan\sqrt{\frac{a-b}{b(a+1)}}=\frac2\pi\phi_2(a,b).\tag3 \end{align}$$ The next integral is also manageable: $$\begin{align} \int_1^\infty f'(z)dz&=\int_1^\infty\int_0^1\frac{dt}{(t^2+b)(z^2t^2+az^2+1)}dz\\ &=\int_1^\infty\int_0^1\frac{1}{1+(a-b)z^2}\left(\frac{1}{t^2+b}-\frac{1}{t^2+a+1/z^2}\right)dtdz\\ &=\int_1^\infty\frac{1}{1+(a-b)z^2}\left(\phi_1(b)-\phi_1(a+1/z^2)\right)dz\\ &=\phi_1(b)\phi_1(a-b)-\int_1^\infty \frac{\arctan(1/\sqrt{a+1/z^2})}{(1+(a-b)z^2)\sqrt{a+1/z^2}}dz\\ &=\phi_1(b)\phi_1(a-b)-\int_0^1 \frac{\arctan(1/\sqrt{t^2+a})}{(t^2+(a-b))\sqrt{t^2+a}}dt\qquad [z\mapsto 1/t]\\ &=\phi_1(b)\phi_1(a-b)-\int_0^1 \frac{\tfrac\pi2-\arctan\sqrt{t^2+a}}{(t^2+(a-b))\sqrt{t^2+a}}dt\\ &=\phi_1(b)\phi_1(a-b)-\frac\pi2\int_0^1 \frac{dt}{(t^2+(a-b))\sqrt{t^2+a}}+\int_0^1 \frac{\arctan\sqrt{t^2+a}}{(t^2+(a-b))\sqrt{t^2+a}}dt\\ &=\phi_1(b)\phi_1(a-b)-\phi_2(a,a-b)+J(a,a-b).\tag4 \end{align}$$ Then from $(3)$ and $(4)$, $$J(a,b)=\phi_2(a,b)-\phi_1(b)\phi_1(a-b)+\phi_2(a,a-b)-J(a,a-b),$$ as desired.

Is there some way to find a closed form for $J$?


EDIT:

One may use the series $$g(z)=\frac{\arctan\sqrt z}{\sqrt z}=\sum_{n\ge1}\frac{(-1)^n}{2n-1}\left(\frac{1}{z^n}-\frac{2}{\sqrt z}\right)\qquad z\ge1$$ to get $$J(a,b)=-\phi_2(a,b)+\sum_{n\ge1}\frac{(-1)^n}{2n-1}\int_0^1\frac{dt}{(t^2+b)(t^2+a)^n},$$ where $\phi_2$ is defined above. The integral $\int_0^1\frac{dt}{(t^2+b)(t^2+a)^n}$ is really not nice though.

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    $\begingroup$ In particular, Iridescent's answer there contains a special value of $$J(2,1+ \frac{2}{\sqrt{5}}) = \sqrt{5} \left(\frac{71 \pi ^2}{3600}+\frac{1}{3} \pi \tan ^{-1}\left(\sqrt{\frac{1}{3} \left(9-4 \sqrt{5}\right)}\right)-\frac{1}{6} \pi \tan ^{-1}\left(\sqrt{27-12 \sqrt{5}}\right)\right)$$ As well as similar one for $J(2,1- \frac{2}{\sqrt{5}})$. I am sure you can dig out more special values of $J(a,b)$ using results there. $\endgroup$
    – pisco
    Feb 27 at 16:07
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    $\begingroup$ I am suspicious that there is a known elementary expression for $J(a,b)$. A restrictive two-variable generalization reads $$\int_0^1\frac{\arctan\sqrt{\frac{t^2+a}b}}{(\frac{1+b}at^2+1)\sqrt{\frac{t^2+a}b}}dt\\ =\frac{\pi^2}8+\frac12\left(\arctan^2\frac1{\sqrt{a+b}} -\arctan^2\sqrt{\frac{1+b}a} -\arctan^2\sqrt{\frac{1+a}b}\right) $$ which contains $J(a,\frac a2)$ as a special case, i.e. $b=1$. $\endgroup$
    – Quanto
    Feb 27 at 17:24
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    $\begingroup$ @Quanto That's nice. You are invited to post that as solution. $\endgroup$
    – pisco
    Feb 27 at 18:25
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    $\begingroup$ @pisco - thanks. I might if I could find my notes $\endgroup$
    – Quanto
    Feb 27 at 20:53
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    $\begingroup$ Either way, we still get the highly nontrivial result (for $0<b<a$) $$\int_0^1\frac{(2t^2+a)\arctan\sqrt{t^2+a}}{t^4+at^2+b(a-b)}\frac{dt}{\sqrt{t^2+a}}=\phi_1(a,b)-\phi_2(b)\phi_2(a-b)+\phi_1(a,a-b),$$ where $\phi_1(a,b)=\frac{\pi}{2\sqrt{b(a-b)}}\arctan\sqrt{\frac{a-b}{b(a+1)}}$ and $\phi_2(x)=\frac{1}{\sqrt x}\arctan\frac{1}{\sqrt x}$. $\endgroup$
    – clathratus
    Mar 6 at 18:21

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