# An integral and a mysterious recursive sequence

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 !

• some solutions math.stackexchange.com/questions/2876586/… Commented Apr 18, 2021 at 14:42
• @Svyatoslav Thanks for the link ! The given recursion formula is the one I used for my computation, but they don't mention the "closed" form in term of the coefficients $C(n,k)$, which is weird since the formula comes easily when you have the recursive relation. Commented Apr 18, 2021 at 14:49
• Since you're after a source: this integral and its recurrence relation appear in page 182 of Companion to Concrete Mathematics: volume I by Z. A. Melzak. Commented Apr 25, 2021 at 5:02

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.

• Great ! But I am a little bit startled : neither the OEIS page nor the Comtet book mention the recursive relation for the $C(n,k)$... They mention another recursive, but far more complicated, relation... Commented Apr 19, 2021 at 6:32
• Great idea, the generating function ! I will have a look to this method. Thanks ! Commented Apr 22, 2021 at 21:00

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 $$c_{n,0}=c_{n,n-1}=\frac{(2n-3)!!}{(2n-2)!!}$$ $$c_{n,k} = \frac{2n-3}{2(n-1)}(c_{n-1,k} +c_{n-1,k-1}) - \frac{n-2}{n-1}c_{n-2,k-1}$$

• Thanks, this is a clever way of solving this integral +1 ! Actually, this is the recursive relation between the coefficients that puzzles me the most. The relation I found appears nowhere and is much simpler than the ones I read... Could you maybe detail which relation you could find with your method ? And do you know if that sequence appears in some other context ? Thanks again ! Commented Apr 22, 2021 at 6:57
• Thanks for the precision ! But again, your recursive formula is quite more complicated than the one I found... Maybe I have been mistaken, but I don't think so... If you have time, could you please check that your coefficients satisfy the relation I wrote in my question ? Thanks ! Commented Apr 22, 2021 at 19:25
• @Henry - I could not quite figure out your recursion. Take $C(2,0)=-C(1,-1)+3C(1,0)$, which does not make sense. Commented Apr 22, 2021 at 20:38
• Sorry, I should have precised that the recursion formula works for $n \geq 2$ and $k\geq 1$. That's why I gave as initial conditions the values of the $C(n,0)$'s. This is exactly the kind of relation you have for binomials, where the $(n,k)$-th coefficient is related to the $(n-1,k-1)$-th and the $(n-1,k)$-th. Does it make more sense ? Commented Apr 22, 2021 at 20:55
• @Henry - Makes sense. I verified that both recursive formuli produce the same coefficients up to $n=5$. So, they must be equivalent. Commented Apr 22, 2021 at 21:35

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.

• Great ! I am a little bit relieved that the integral cannot be solved explicitely in term of usual functions... I was afraid of having missed something simple. Do you know if these integrals appear in some context ? And something about the recursive sequence that intrigues me ? Thanks a lot ! Commented Apr 19, 2021 at 6:35
• @Henry. I am also stuck with the recursive sequence. Commented Apr 19, 2021 at 6:38
• Ok, thanks for your time ! Let's wait for - hopefully - other answers... Commented Apr 19, 2021 at 6:39

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

$$J_{n+1}(p)=J_n(p)+\frac{p-1}n J_n’(p)\tag 1$$ 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}

• Thanks for this answer ! Yes, this is basically the method I used first, except that I udes partial differentiation w.r.t. $a$ and $b$, rather than using a single variable as you did here (which is indeed a good way to simplify a little bit the calculation !) Commented Apr 24, 2021 at 20:05

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.

• Thanks for your answer ! Actually, I should have precised that the initial values of $C(n,k)$ are given by $C(1,0)=C(1,1)=1$. So the coefficients are a little bit different... if I have not been mistaken in my calculations ! I will edit my question to make the first values of $C(n,k)$ appear. Commented Apr 18, 2021 at 17:23
• In this case, I think you should also specify $C(n,0)$ for $n\gt1$ as an additional initial condition. Commented Apr 18, 2021 at 18:58
• Yes, indeed. Actually, they must satisfy $C(n,0)=C(n,n)=\frac{(2n)!}{2^nn!}$, I add this to the question ! Thanks for your comment ! Commented Apr 18, 2021 at 20:52