Is it possible to integrate $\frac{ \tan ^{-1}(t)}{t^{2n}\,\sqrt{t^2-1}}$

Question

Working on this question, I faced the problem of computing:

$$I_n=\int_1^a \frac{ \tan ^{-1}(t)}{t^{2n}\,\sqrt{t^2-1}}\,dt$$

For a given value of $n \geq 1$, Mathematica does not face any problem and the results are quite simple.

For example, if $a=\sqrt 3$ as in the linked question $$I_{5}=\frac{8346321-4992192 \sqrt{2}+3246848 \sqrt{6}}{58786560}\pi+$$ $$\frac{7867 \sqrt{2}-309123 \cot ^{-1}\left(\sqrt{2}\right)}{1088640}$$

My question is : Is it possible to find the antiderivative even in terms of special functions ?

Answer 1

Although I believe there is no closed form for $I_n$ for any values of $n\ge 0$, in the following post, we will provide a recurrence relation for $I_n$ in terms of well-known, closed-form functions.

We proceed by applying IBP to $I_n$: $$ I_n { = \int_1^a\frac{\tan^{-1}t}{t^{2n}\sqrt{t^2-1}}dt = \int_1^a\frac{\tan^{-1}t}{t^{2n+1}}\frac{tdt}{\sqrt{t^2-1}} = \int_{t=1}^{t=a}\frac{\tan^{-1}t}{t^{2n+1}}d(\sqrt{t^2-1}) \\= \frac{\sqrt{t^2-1}\tan^{-1}t}{t^{2n+1}}\Bigg|_1^a +(2n+1) \int_1^a\frac{\sqrt{t^2-1}\tan^{-1}t}{t^{2n+2}}dt - \int_1^a\frac{\sqrt{t^2-1}}{t^{2n+1}(1+t^2)}dt \\= \frac{\sqrt{a^2-1}\tan^{-1}a}{a^{2n+1}} +(2n+1) \int_1^a\frac{(t^2-1)\tan^{-1}t}{t^{2n+2}\sqrt{t^2-1}}dt \\- \int_0^{\cos^{-1}\frac{1}{a}}\frac{\cos^{2n}\theta-\cos^{2n+2}\theta }{1+\cos^2\theta} d\theta \\= \frac{\sqrt{a^2-1}\tan^{-1}a}{a^{2n+1}} +(2n+1)\left( \int_1^a\frac{\tan^{-1}t}{t^{2n}\sqrt{t^2-1}}dt - \int_1^a\frac{\tan^{-1}t}{t^{2n+2}\sqrt{t^2-1}}dt \right) \\- \int_0^{\cos^{-1}\frac{1}{a}} \frac{2(-1)^n}{1+\cos^2\theta} -\cos^{2n}\theta+\sum_{k=0}^{n-1}2(-1)^{n-k-1}\cos^{2k}\theta d\theta \\= \frac{\sqrt{a^2-1}\tan^{-1}a}{a^{2n+1}} -\sqrt2(-1)^n\tan^{-1}\sqrt{\frac{a^2-1}{2}} \\+(2n+1)\left( I_n - I_{n+1} \right)+J_n+\sum_{k=0}^{n-1}2(-1)^{n-k}J_k, } $$ where $J_n=\int_0^{\cos^{-1}\frac{1}{a}}\cos^{2n}\theta d\theta$. The following recurrence is then deduced: $$ I_{n+1}{= \frac{\sqrt{a^2-1}\tan^{-1}a}{(2n+1)a^{2n+1}} -\frac{\sqrt2(-1)^n\tan^{-1}\sqrt{\frac{a^2-1}{2}}}{2n+1} \\+\frac{2nI_n}{2n+1}+\frac{J_n}{2n+1}+\sum_{k=0}^{n-1}\frac{2(-1)^{n-k}}{2n+1}J_k }. $$ For $J_n$, we can write $$ J_n {= \int_0^{\cos^{-1}\frac{1}{a}}\cos^{2n}\theta d\theta = \int_0^{\cos^{-1}\frac{1}{a}}\cos^{2n-1}\theta \cos \theta d\theta \\=\cos^{2n-1}\theta \sin \theta|_0^{\cos^{-1}\frac{1}{a}}+(2n-1)\int_0^{\cos^{-1}\frac{1}{a}} \sin^2\theta\cos^{2n-2}\theta \\=\frac{\sqrt{a^2-1}}{a^{2n}}+(2n-1)\int_0^{\cos^{-1}\frac{1}{a}} \cos^{2n-2}-\cos^{2n}\theta d\theta \\=\frac{\sqrt{a^2-1}}{a^{2n}}+(2n-1)J_{n-1}-(2n-1)J_n, } $$ which yields $$ J_n=\frac{\sqrt{a^2-1}}{2na^{2n}}+\frac{2n-1}{2n}J_{n-1}\quad,\quad J_0=\cos^{-1}\frac{1}{a}. $$

Conclusion

The following recurrence is given for $I_n$: $$ I_{n+1} {= \frac{\sqrt{a^2-1}\tan^{-1}a}{(2n+1)a^{2n+1}} -\frac{\sqrt2(-1)^n\tan^{-1}\sqrt{\frac{a^2-1}{2}}}{2n+1} \\+\frac{2nI_n}{2n+1}+\frac{J_n}{2n+1}+\sum_{k=0}^{n-1}\frac{2(-1)^{n-k}}{2n+1}J_k }, $$ where $$ J_n=\frac{\sqrt{a^2-1}}{2na^{2n}}+\frac{2n-1}{2n}J_{n-1}\quad,\quad J_0=\cos^{-1}\frac{1}{a} $$ and $I_0=\int_1^a\frac{\tan^{-1}t}{\sqrt{t^2-1}}dt$.

Answer 2

In terms of an infinite summation, writing $$\frac{1}{\sqrt{t^2-1}}=\sum_{k=0}^\infty (-1)^k \,\binom{-\frac{1}{2}}{k} \,\,t^{-(2 k+1)}$$ we have $$I_n=\sum_{k=0}^\infty (-1)^k \,\binom{-\frac{1}{2}}{k} \,\,\int_1^a t^{-(2 k+2 n+1)}\tan ^{-1}(t)$$ $$J_k=\int_1^a t^{-(2 k+2 n+1)}\tan ^{-1}(t)$$ $$8(k+n)J_k=\pi- \frac{4\,\tan ^{-1}(a)}{ a^{2 (k+n)}}-2 i e^{i \pi (k+n)} B_{-a^2}\left(-k-n+\frac{1}{2},0\right)+$$ $$\Big(H_{-\frac{2 k+2 n+3}{4} }-H_{-\frac{2 k+2 n+1}{4} }\Big)$$ and we can use the recurrence relation for the incomplete beta function.

Answer 3

Another infinite summation $$\frac{\tan ^{-1}(t)}{\sqrt{t^2-1}}=\sum_{k=0}^\infty (-1)^k\, \alpha_k\,(t-1)^{\frac{2k-1}{2} }$$

$$J_k=\int_1^a t^{-2n}\,(t-1)^{\frac{2k-1}{2} }\,dt$$ $$J_k=\frac{\Gamma \left(k+\frac{1}{2}\right) \Gamma \left(2 n-k-\frac{1}{2}\right)}{\Gamma (2 n)}-B_{\frac{1}{a}}\left(2 n-k-\frac{1}{2},k+\frac{1}{2}\right)$$

The coefficients of the expansion are

$$\alpha_k=\frac 1{\sqrt 2}\Bigg(\frac{\pi }{2^{p+2}\Gamma (p+1)}\left(\frac{1}{2}\right)_p-\beta_k\Bigg)$$ where the $\beta_k$ form the (unknown ?) sequence $$\left\{0,\frac{1}{2},\frac{3}{8},\frac{37}{192},\frac{49}{768},\frac {23}{20480},-\frac{1287}{81920},-\frac{52167}{4587520},-\frac{1 0355}{3670016},\cdots\right\}$$

Answer 4

Not an answer just an approach

Using the well-know formula $\arctan\left(x\right)+\operatorname{arccot}\left(x\right)=\pi/2$ and $n\geq 2$ an integer ,$x>1$:

$$\int_{ }^{ }\frac{1}{x^{2n}}\cdot\frac{\arctan\left(x\right)+\operatorname{arccot}\left(x\right)}{\sqrt{x^{2}-1}}dx=\frac{\sqrt{x^{2}-1}\left(\sum_{n=0}^{M}a_{n}x^{2n}\right)\left(\arctan\left(x\right)+\operatorname{arccot}\left(x\right)\right)}{C_{n}x^{2M+1}}$$

For some positive constantes and coefficients . Then for $0<x<1$

$$\arctan\left(x\right)-\operatorname{arccot}\left(x\right)=2x-\frac{2}{3}x^{3}+\frac{2}{5}x^{5}+\cdot\cdot\cdot$$

Then using the recurrence relation in the other answer we can use Ramanujan's master theorem setting $x\to 1/x$.

A conjecture :

Let $0.27<x<0.3$ and define for $q,c_i\in(0,\infty)$ :

$$f\left(x\right)=\frac{\pi}{4}\cdot\frac{x-1}{x}\left(1-\frac{x-1}{8\sqrt{x^{2}-1}}\left(2+\sum_{n=3}^{P}\left(u+\frac{1}{n}\right)^{-c_{n}x}\right)\right)$$

Then there exists positive constants $k,m,\alpha>0$ such that :

$$\lim_{P\to \infty}f(x) \implies \arctan\left(x\right)-f\left(\frac{1}{x}\right)-\left(kx^{\alpha}-m\right)=0$$

Then there is exponential integral and a possible telescoping in $I_0$.

On the other hand it's like a Troyan's horse using :

Let $x,y\in[0,1)$ then we have :

$$\arctan\left(x\right)+\arctan\left(y\right)-\arctan\left(\frac{x+y}{1-xy}\right)=0$$