An integral and a mysterious recursive sequence

Question

Today I tried to solve the integral $$I_n= \int_0^{\frac{\pi}{2}} \frac{\mathrm{dt}}{(a \cos^2(t)+ b \sin^2(t))^n} \quad \quad \quad \quad a,b > 0$$

After some dirty calculations, I obtained that $$I_n = \frac{\pi}{2^n(n-1)!\sqrt{ab}} \sum_{k=0}^{n-1} \frac{C(n-1,k)}{a^kb^{n-1-k}}$$

where the $C(n,k)$ are defined for $n \geq 1$ and $0 \leq k \leq n$ and satisfy the "Pascal-like" recursive relation $$C(n,k)=(2k-1) \times C(n-1,k-1)+(2n-2k-1) \times C(n-1,k) \quad \quad n\geq 2, \text{ } k \geq 1$$ with the initial conditions $$C(1,0)=C(1,1)=1 \quad \quad \text{and} \quad \quad C(n,0)=C(n,n)=\frac{(2n)!}{2^nn!} \text{ for all } n$$

My question are :

• is that a "well-known" sequence of coefficients ? Do you know if they appear in some other problems ?
• are also these integrals well-known ? Do you maybe know an elegant way to solve them ?

In a summary, can you tell me anything you know about these integrals and this recursive sequence !

Thanks a lot !

EDIT : As @ReneGy pointed out, this sequence is apparently registered is OEIS as A059366. But OEIS does not mention the recursive relation I gave... I am still taking anything you can tell me about this mysterious sequence and these integrals !

Answer 1

Now, the new triangular table for $C(n,k)$

n\k	1	2	3	4	5	6	7
1	1	1
2	3	2	3
3	15	9	9	15
4	105	60	54	60	105
5	945	525	450	450	525	945
6	10395	5670	4725	4500	4725	5670	10395

This is A059366 in the OEIS. Apparently, these numbers were already found for the calculation of $$I_n= \int_0^{\frac{\pi}{2}} \frac{\mathrm{dt}}{(a \cos^2(t)+ b \sin^2(t))^n} \quad \quad \quad \quad a,b > 0,$$see L. Comtet, Advanced Combinatorics, Reidel, 1974, pp. 166-167.

We have:

$$ \sum_{n\ge 0}I_nx^n=\int_0^{\frac{\pi}{2}}\mathrm{dt}\sum_{n\ge 0} \frac{x^n}{(a \cos^2(t)+ b \sin^2(t))^n}= \int_0^{\frac{\pi}{2}}\mathrm{dt}\frac{1}{1-\frac{x}{a \cos^2(t)+ b \sin^2(t)}}$$ $$ \sum_{n\ge 0}I_nx^n= \int_0^{\frac{\pi}{2}}\frac{a \cos^2(t)+ b \sin^2(t)}{a \cos^2(t)+ b \sin^2(t)-x}\mathrm{dt}=\frac{\pi}{2}+\int_0^{\frac{\pi}{2}}\frac{x}{a \cos^2(t)+ b \sin^2(t)-x}\mathrm{dt}$$ $$ \sum_{n\ge 0}I_nx^n=\frac{\pi}{2}+\int_0^{\frac{\pi}{2}}\frac{x}{a-x + (b-x) \tan^2(t)}\mathrm{d\tan{t}}=\frac{\pi}{2}+\int_0^{\infty}\frac{x}{a-x + (b-x) t^2}\mathrm{dt}$$ $$ \sum_{n\ge 0}I_nx^n=\frac{\pi}{2}+\frac{x}{a-x}\int_0^{\infty}\frac{1}{1 + (\frac{\sqrt{b-x}}{\sqrt{a-x}}t)^2}\mathrm{dt}=\frac{\pi}{2}+\frac{x}{\sqrt{(a-x)(b-x)}}\int_0^{\infty}\frac{1}{1 + t^2}\mathrm{dt}$$ $$ \sum_{n\ge 0}I_nx^n=\frac{\pi}{2}\big(1+\frac{x}{\sqrt{(a-x)(b-x)}}\big)$$

then , for $n\gt 0$ $$I_n = \frac{\pi}{2} [x^{n}]\frac{x}{\sqrt{(a-x)(b-x)}}$$ but $$ \sum_{i\ge 0}{2i\choose i}\big(\frac{x}{a}\big)^i=\frac{\sqrt a}{\sqrt{a-4x}}.$$

Then we have $$I_n = \frac{\pi}{2^{2n-1}\sqrt{ab}} \sum_{i+j=n-1} \frac{{2i\choose i}}{a^i}\frac{{2j\choose j}}{b^j}.$$

By identification to the OP formula, this gives
$$C(n,k)= \frac{n!}{2^n}{2k\choose k}{2(n-k)\choose n-k} $$

which is $$C(n,k)= (-2)^n{n\choose k}{-\frac{1}{2}\choose k} {-\frac{1}{2}\choose n-k}$$ which is given in the above OEIS link (with a typo).

It should be not so difficult to verify that these closed formulae for $C(n,k)$ do satisfy the above recursion.

Answer 2

Note that \begin{align} & \left( \frac{(b-a)\sin t\cos t}{(a \cos^2 t+ b \sin^2 t)^{n-1}} \right)’\\ = & -\frac{{2(n-1)ab}}{(a \cos^2 t+ b \sin^2 t)^n} + \frac{{(2n-3)(a+b)}}{(a \cos^2 t+ b \sin^2 t)^{n-1}} - \frac{{2(n-2)}}{(a \cos^2 t+ b \sin^2 t)^{n-2}} \end{align} Integrate both sides to establish $$(n-1)I_n = (n-\frac32)\left(\frac1a+\frac1b\right)I_{n-1}-\frac{n-2}{ab}I_{n-2} $$ with $I_1 =\frac\pi{2\sqrt{ab}}$. Then

\begin{align} & I_2 = \frac12 I_1\left(\frac1a+\frac1b\right) \\ & I_3 = \frac18I_1\left(\frac3{a^2}+\frac2{ab}+\frac3{b^2}\right)\\ &\>\cdots\\ & I_n = I_1 \left(\frac{c_{n,0}}{a^{n-1} }+ \frac{c_{n,1}}{a^{n-2}b} + \cdots + \frac{c_{n,n-1}}{b^{n-1} }\right) \end{align} where the coefficients satisfy the recursive relationship $c_{n,0}=c_{n,n-1}=\frac{(2n-3)!!}{(2n-2)!!}$ $$c_{n,k} = \frac{2n-3}{2(n-1)}(c_{n-1,k} +c_{n-1,k-1}) - \frac{n-2}{n-1}c_{n-2,k-1}$$

Answer 3

There is an antiderivative for $$J_n= \int \frac{\mathrm{dt}}{\big[a \cos^2(t)+ b \sin^2(t)\big]^n} $$ (have a look here) in terms of the Appell hypergeometric function of two variables.

Using the integration bounds, $$I_n=-\frac{\sqrt{\pi } \sqrt{1-\frac{a}{b}} \sqrt{1-\frac{b}{a}} \,\Gamma (2-n) }{2 (n-1) (a-b) (ab)^n \,\Gamma \left(\frac{3}{2}-n\right)} \,A$$ where $$A=a^n b\,\, _2F_1\left(\frac{1}{2},1-n;\frac{3-2n}{2};\frac{b}{a}\right)+a b^n \,\, _2F_1\left(\frac{1}{2},1-n;\frac{3-2n}{2};\frac{a}{b}\right)$$ where appears the gaussian hypergeometric function.

Answer 4

Start with

$$J_n(p)=\frac2\pi \int_0^{\frac{\pi}{2}} \frac{{dt}}{(p \cos^2t+ \sin^2t)^n}$$

$$J_{n+1}(p)=J_n(p)+\frac{p-1}n J_n’(p)\tag 1 $$ With $J_1(p)= \frac1{p^{1/2}} $ \begin{align} & J_2 (p)=\frac12\left(\frac1{p^{1/2}}+\frac1{p^{3/2}}\right)\\ & J_3 (p)= \frac18\left(\frac3{p^{1/2}}+\frac2{p^{3/2}}+\frac3{p^{5/2}}\right)\\ & J_4 (p)= \frac1{16}\left(\frac5{p^{1/2}}+\frac3{p^{3/2}}+\frac3{p^{5/2}}+\frac5{p^{7/2}}\right)\\ & \cdots\>\cdots\\ & J_{n}(p) = \frac{c_{n-1,0} }{p^{1/2}}+ \frac{c_{n-1,1} }{p^{3/2}}+\cdots + \frac{c_{n-1,n-1} }{p^{(2n-1)/2}}\\ & J_{n+1}(p) = \frac{c_{n,0} }{p^{1/2}}+ \frac{c_{n,1} }{p^{3/2}}+\cdots + \frac{c_{n,n-1} }{p^{(2n-1)/2}}+\frac{c_{n,n} }{p^{(2n+1)/2}} \end{align}

with the coefficients satisfying the recursion $c_{0,0}=1,\>c_{n,0}= \frac{2n-1}{2n}c_{n-1,0}$ $$c_{n,k}=\frac{k-n-1}kc_{n,k-1} +\frac nk c_{n-1,k-1}, \>\>\>\>\>k\ne 0$$

Then

\begin{align} I_{n+1} &= \int_0^{\frac{\pi}{2}} \frac{{dt}}{(a \cos^2t+ b\sin^2t)^{n+1}} =\frac\pi2 \frac1{b^{n+1}} J_{n+1}\left(\frac ab \right)\\ &= \frac\pi2\left(\frac{c_{n,0} }{a^{1/2}b^{(2n+1)/2}}+ \frac{c_{n,1} }{a^{3/2}b^{(2n-1)/2}}+\cdots + \frac{c_{n,n} }{a^{(2n+1)/2}b^{1/2}}\right) \end{align}

Answer 5

Not an answer, but too long for comment: here is a triangular table for $C(n,k)$

n\k	0	1	2	3	4	5	6
0	1
1	0	1
2	0	1	3
3	0	3	6	15
4	0	15	27	45	105
5	0	105	180	270	420	945
6	0	945	1575	2250	3150	4725	10395

It does not seem to be in the OEIS.