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When I evaluate the following definite integral, $$I=\int_{0}^{\pi/2}\frac{\mathrm dx}{1+\sin^4x},\tag 1$$ I have two methods: the first one: let $u=\cot x$, then $$I=\int_{0}^{\infty}\frac{u^2+1}{u^4+2u^2+2}\mathrm du=\frac{\sqrt{1+\sqrt{2}}}{4}\pi,$$ this is a integral about rational function who always has an antiderivative (it is not easy to get the antiderivative).

The second method is: $$I=\int_{0}^{\pi/2}\sum_{n=0}^{\infty}(-1)^n\sin^{4n}x\ \mathrm dx =\sum_{n=0}^{\infty}(-1)^n\int_{0}^{\pi/2}\sin^{4n}x\ \mathrm dx =\frac{\pi}{2}\sum_{n=0}^{\infty}(-1)^n\frac{(4n-1)!!}{(4n)!!}.$$ The series $\sum_{n=0}^{\infty}(-1)^n\frac{(4n-1)!!}{(4n)!!}$ is convergent, but I have no method to evaluate the sum of this series directly.

Combining above two methods, we can get $$\sum_{n=0}^{\infty}(-1)^n\frac{(4n-1)!!}{(4n)!!}=\frac{\sqrt{1+\sqrt{2}}}{2}.\tag 2$$

What I concern most is how to evaluate series $(2)$ directly and evaluate the general integral $$I_n=\int_{0}^{\pi/2}\frac{\mathrm dx}{1+\sin^nx},\quad n=1,2,3,4,\cdots.$$ Does $I_n$ have closed form? For $n=1,2,4$, we have $$I_1=1,\quad I_2=\frac{\pi}{2\sqrt2},\quad I_3=?,\quad I_4=\frac{\sqrt{1+\sqrt{2}}}{4}\pi.$$

Riemann
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  • I doubt there is any closed-form expression for $I_n$. CAS shows $I_3$ has a very unwieldy expression, which I can't even put in this comment box. You can check it out on SageMath. – Sahaj Jan 20 '24 at 06:17
  • @ user12030145 Thanks you for your reference! – Riemann Jan 20 '24 at 10:35
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  • @ Random Variable Thank you very much! – Riemann Jan 21 '24 at 09:38
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    $$\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n} \left( 4,n-1 \right) !!}{ \left( 4,n \right) !!}}={\frac {\sin \left( {\frac {3,\pi}{8}} \right) }{2^{1/4}}} \ \sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{n} \left( 4,n-1 \right) !!{x}^{n}}{ \left( 4,n \right) !!}}={\frac {1}{\sqrt2} \sqrt {1+\sqrt {1+x}}{\frac {1}{\sqrt {1+x}}}}$$ – GEdgar Jan 26 '24 at 02:35

3 Answers3

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For the summation $$f(x)=\sum_{n=0}^{\infty}(-1)^n\frac{(4n-1)!!}{(4n)!!}\,x^n=\frac{1}{\sqrt{\pi }}\sum_{n=0}^{\infty}(-1)^n\, \frac{\Gamma \left(2 n+\frac{1}{2}\right)}{\Gamma (2 n+1)}\,x^n$$ $$f(x)=\frac{\sqrt{1+\sqrt{x+1}}}{ \sqrt{2(x+1)}}$$

For the integrals

With regard to $I_3$, the only solution I found is based on the tangent half-angle substitution $$I_3=\int_0^1 \frac{2\left(t^2+1\right)^2}{(t+1)^2 \left(t^4-2 t^3+6 t^2-2t+1\right)}\,dt$$

Factoring the quartic and using partial fraction decomposition, the integrand is

$$\frac{2}{3 (t+1)^2}+\frac{2}{3 t^2-3 \left(1+i \sqrt{3}\right) t+3}+\frac{2}{3 t^2-3\left(1-i \sqrt{3}\right) t+3}$$

So, three simple integrals. Using the bounds

$$\color{blue}{I_3=\frac 13-\frac{1}{3\sqrt[4]{12}}\Big(\left(\sqrt{3}-1\right)+i (\sqrt{3}+1) \Big)\tanh ^{-1}\left((1+i)\frac{ \sqrt[4]{3}}{\sqrt{2}}\right)+}$$ $$\color{blue}{\frac{1}{3 \sqrt[4]{12}}\Big(\left(\sqrt{3}+1\right)+i (\sqrt{3}-1) \Big)\tan ^{-1}\left((1+i)\frac{ \sqrt[4]{3}}{\sqrt{2}}\right)}$$ which is probably the most compact form. Expanding the complex number would lead to four logarithms and four arctangents.

The final result is

$$\color{red}{I_3= \frac 13+\pi \sqrt{\frac{1}{9}+\frac{2}{9 \sqrt{3}}}-\frac{1}{3} \sqrt{1+\frac{2}{\sqrt{3}}} \tan ^{-1}\left(\sqrt{3+2 \sqrt{3}}\right)+}$$ $$\color{red}{\frac{\sqrt{3} \log \left(1+\sqrt{3}-\sqrt{3+2 \sqrt{3}}\right)+\log \left(1+\sqrt{3}+\sqrt{3+2 \sqrt{3}}\right)}{6 \sqrt[4]{12}}}$$

From a formal point of view, the same could be done for any $n$ with a major problem because of the denominator in $$I_n= \int_0^1 \frac{2\left(t^2+1\right)^{n-1}}{2^n t^n+\left(t^2+1\right)^n}\,dt$$

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Let us try not to use the (brilliant, yet a bit cumbersome) solution given 4 years ago, mentioned in the question comment by Random Variable and find a special function solution instead. Note that what a "closed-form" solution refers to is subject to historical variations: elliptic functions did not belong to the set of closed-form primitives before they were developed in the mid-19th century. Similarly, Fox-Wright and Fox-H functions may not "look like" closed-form primitives for an undergraduate student and will look so at a higher level. I am choosing the latter convention. Note that implementation in standard CAS is also to be considered. Mathematica, for one, has the function $\texttt{FoxH}$. Considering now

$$I_n = \int_0^{\frac{\pi}{2}} \frac{dx}{1+\sin^n(x)} = \sum_{m=0}^\infty (G_{m,n} - H_{m,n})$$

in which $G_{m,n} = \int_0^{\frac{\pi}{2}} \sin^{2mn}(x)\,dx, H_{m,n} = \int_0^{\frac{\pi}{2}} \sin^{(2m+1)n}(x)\,dx$

(Edit)

Note that the sums cannot be separated out (as in the alternate harmonic series), as they individually diverge:

$$ H_{m,n} \sim \sqrt{\frac{e\pi}{2(2m+1)n}}, \quad G_{m,n} \sim \sqrt{\frac{e}{\pi mn}}$$

By Gradshtein et al. [2010, eq. 3.621.3 p. 397],

$$G_{m,n} = \frac{(2mn-1)!!}{(2mn)!!} \frac{\pi}{2} = \frac{\Gamma(2mn)}{2^{2mn}\Gamma(mn)\Gamma(mn+1)}\pi,$$

for $m > 0$, with the convention $(-1)!!=1$. This formula may be simplified by way of Legendre's duplication formula

$$ \Gamma(mn)\, \Gamma(mn+\frac{1}{2}) = \Gamma(2mn) \; 2^{-2mn+1} \sqrt{\pi}$$

whence:

$$G_{m,n} = \frac{\sqrt{\pi}\Gamma\left(mn+\frac{1}{2}\right)}{2 \Gamma(mn+1)},$$

which has the advantage of being also true for $m=0$, and by eq. 3.621.4,

$$H_{m,k} = \frac{((2m+1)n-1)!!}{((2m+1)n)!!} \delta$$ \begin{equation} H_{m,k} = \begin{cases}\frac{\sqrt{\pi} \Gamma((m+\frac{1}{2})n + \frac{1}{2})}{2 \Gamma((m+\frac{1}{2})n + 1)}\qquad \text{if $n$ is even}\\ 2^{n-1}\frac{2^{2mn}\Gamma(mn+\frac{n-1}{2})^2}{\Gamma(2mn+n+1)} \qquad \qquad \qquad \text{if $n$ is odd} \end{cases} \end{equation}

for $m \geq 0$, in which $\delta= \frac{\pi}{2}$ if $n$ is even and $\delta=1$ if $n$ is odd.

As the sums of $(H_{m,n})_{m\geq 0}$ and $(G_{m,n})_{m\geq 0}$ individually diverge but converge if $(I_{m,n}=G_{m,n}-H_{m,n})_{m\geq 0}$ is considered, and the form of $H_{m,n}$ is very different from that of $G_{m,n}$ when $n$ is odd, we will be able to find a solution only for even values of $n$ using the special function method outlined below. Further developments should derive $I_n$ for odd values of $n$ from the preceding even values.

Considering therefore sums with even $n$, we can rewrite it as a Fox-Wright function (and not as a q-hypergeometric series as I first thought):

$$\sum_{m=0}^\infty I_{m,n}=\pi \, \left(\frac{1}{2} + \sum_{p=1}^\infty \frac{\Gamma(pn)}{2^{pn} \Gamma(\frac{p}{2}n)\Gamma(\frac{p}{2}n+1)}(-1)^p\right) = \,\pi\, {_2}\Psi_{2}\left[\begin{matrix}(0,n)&(1,1)\\ (0,\frac{n}{2})&(1,\frac{n}{2})\end{matrix}; -2^{-n}\right]$$

There is some basic algebra in the above formulae, which I still have not thoroughly checked (I will come back for edits if need be).

Hence the closed form expressed for $n$ even.

Note: Convergence conditions of the Fox-Wright function have not been checked, but my take is that the convergence of Fox-Wright functions is ensured by the existence of the integrals. For a closer view on convergence you may find it useful to check Mathai et al.'s The H function, Springer 2010, page 30.

To implement it using the Mathematica $\texttt{FoxH}$ function, simply use the fact that

$${_2}\Psi_{2}\left[\begin{matrix}(0,n)&(1,1)\\ (0,\frac{n}{2})&(1,\frac{n}{2})\end{matrix}; -2^{-n}\right] = H^{1,2}_{2,3}\left[\begin{matrix}(1,n)&(0,1)\\ (0,1) & (1,\frac{n}{2})&(0,\frac{n}{2})\end{matrix}; 2^{-n}\right]$$

user12030145
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Not a complete solution but maybe some progress for even indexed $I_n$. By expanding the denominator as a geometric series (as you did for $I_4$), recognising the Wallis integral and using a root of unity filter on the resulting sum, we can find that $$I_{2n} = \frac{\pi}{2n}\sum_{z \in A_n}\text{pv}\frac{1}{\sqrt{1-z}}$$ where $A_n$ is the set of nth roots of $-1$, i.e, $A_n=\{z \in \mathbb{C}:z^n=-1 \}$. I've tried simplifying this sum further, either through the definition of the principal square root or by trying to exploit that the sum is symmetric in terms of roots of a polynomial but I've had little success.