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This integral cannot be solved with any standard function

$$I=\int_a^b \frac{\sin u}{(c-u-\sin u)^{2/3}}du$$

$c,a,b\in \mathbb{R}^+\;\;,\;\;c>u+\sin u \, \forall\;u\in[a,b]$

It is possible to solve this integral using any Special Functions ( Elliptic Integrals, Hypergeometric Functions, Gamma function etc..)?

C.C.12
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  • Mathematica balks at it. You might try differentiation under the integral sign, but there's a lot of different avenues there. – Adrian Keister Oct 18 '19 at 16:42
  • If you mean integration by parts, yes, it doesn't change much – C.C.12 Oct 18 '19 at 17:14
  • What is meant by "differentiation under the integral" is the use of the formula $$\frac{d}{dt}\int_{x_1}^{x_2}f(t,x)dx=\int_{x_1}^{x_2}\frac{\partial}{\partial t}f(t,x)dx$$ where $x_1,x_2$ are constants in terms of $t$. – clathratus Oct 18 '19 at 17:20
  • What would it be the use of "differentiation under the integral" for solving this integral? The function in the integral is only function of u – C.C.12 Oct 18 '19 at 17:31
  • @C.C.12 One often introduces a new parameter with respect to which to differentiate. – MathIsFun7225 Oct 18 '19 at 22:18
  • @MathlsFun7225 I'm sorry, but I don't think I understood. Could you give me an example please? Even on this specific case maybe – C.C.12 Oct 19 '19 at 14:33
  • @C.C.12 see my answer here – mathworker21 Oct 20 '19 at 21:36
  • @ mathworker21 Ah ok, the Feynman method ! But in this case,doing the calculations, it doesn't seem to me that it solves anything, or i am wrong? – C.C.12 Oct 21 '19 at 17:50
  • I don;t think it's possible to express this in closed-form... – Infiniticism Oct 26 '19 at 09:17
  • "The function in the integral is only function of u". False. The integrand is a function of $u$ and $c$. – Eric Towers Jan 07 '20 at 17:32

2 Answers2

2

$\int_a^b\dfrac{\sin u}{(c-u-\sin u)^\frac{2}{3}}~du$

$=\int_a^b\dfrac{\sin u}{c^\frac{2}{3}\left(1-\dfrac{u+\sin u}{c}\right)^\frac{2}{3}}~du$

$=\int_a^b\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{2}{3}\right)_n(u+\sin u)^n\sin u}{c^{n+\frac{2}{3}}n!}~du$ (according to http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric1F0/02)

$=\int_a^b\sum\limits_{n=0}^\infty\sum\limits_{m=0}^n\dfrac{\left(\dfrac{2}{3}\right)_nu^m\sin^{n-m+1}u}{c^{n+\frac{2}{3}}m!(n-m)!}~du$

$=\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=m}^\infty\dfrac{\left(\dfrac{2}{3}\right)_nu^m\sin^{n-m+1}u}{c^{n+\frac{2}{3}}m!(n-m)!}~du$

$=\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{2}{3}\right)_{m+n}u^m\sin^{n+1}u}{c^{m+n+\frac{2}{3}}m!n!}~du$

$=\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{2}{3}\right)_{m+2n}u^m\sin^{2n+1}u}{c^{m+2n+\frac{2}{3}}m!(2n)!}~du+\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{2}{3}\right)_{m+2n+1}u^m\sin^{2n+2}u}{c^{m+2n+\frac{5}{3}}m!(2n+1)!}~du$

$=\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^n\dfrac{\Gamma\left(m+2n+\dfrac{2}{3}\right)C_{n+p+1}^{2n+1}u^m\sin((2p+1)u)}{\Gamma\left(\dfrac{2}{3}\right)4^nc^{m+2n+\frac{2}{3}}m!(2n)!}~du+\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\Gamma\left(m+2n+\dfrac{5}{3}\right)C_{n+1}^{2n+2}u^m}{\Gamma\left(\dfrac{2}{3}\right)4^{n+1}c^{m+2n+\frac{5}{3}}m!(2n+1)!}~du-\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^n\dfrac{\Gamma\left(m+2n+\dfrac{5}{3}\right)(-1)^pC_{n+p+2}^{2n+2}u^m\cos((2p+2)u)}{\Gamma\left(\dfrac{2}{3}\right)2^{2n+1}c^{m+2n+\frac{5}{3}}m!(2n+1)!}~du$

(according to https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae)

$=\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\Gamma\left(m+2n+\dfrac{5}{3}\right)u^m}{\Gamma\left(\dfrac{2}{3}\right)2^{2n+1}c^{m+2n+\frac{5}{3}}m!n!(n+1)!}~du-\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^n\dfrac{\Gamma\left(m+2n+\dfrac{5}{3}\right)(-1)^p(n+1)u^m\cos((2p+2)u)}{\Gamma\left(\dfrac{2}{3}\right)4^nc^{m+2n+\frac{5}{3}}m!(n-p)!(n+p+2)!}~du+\int_a^b\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^n\dfrac{\Gamma\left(m+2n+\dfrac{2}{3}\right)(2n+1)u^m\sin((2p+1)u)}{\Gamma\left(\dfrac{2}{3}\right)4^nc^{m+2n+\frac{2}{3}}m!(n-p)!(n+p+1)!}~du$

$=\left[\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{\Gamma\left(m+2n+\dfrac{5}{3}\right)u^{m+1}}{\Gamma\left(\dfrac{2}{3}\right)2^{2n+1}c^{m+2n+\frac{5}{3}}(m+1)!n!(n+1)!}\right]_a^b-\left[\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^n\sum\limits_{q=0}^m\dfrac{\Gamma\left(m+2n+\dfrac{5}{3}\right)(-1)^p(n+1)u^{m-q}\sin\left((2p+2)u+\dfrac{q\pi}{2}\right)}{\Gamma\left(\dfrac{2}{3}\right)4^nc^{m+2n+\frac{5}{3}}(2p+2)^{q+1}(m-q)!(n-p)!(n+p+2)!}\right]_a^b-\left[\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{p=0}^n\sum\limits_{q=0}^m\dfrac{\Gamma\left(m+2n+\dfrac{2}{3}\right)(2n+1)u^{m-q}\cos\left((2p+1)u+\dfrac{q\pi}{2}\right)}{\Gamma\left(\dfrac{2}{3}\right)4^nc^{m+2n+\frac{2}{3}}(2p+1)^{q+1}(m-q)!(n-p)!(n+p+1)!}\right]_a^b$

(according to https://en.wikipedia.org/wiki/List_of_integrals_of_trigonometric_functions)

Harry Peter
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  • This however does not solve the problem. The solution is still expressed in integral temrines and not in terms of any special function or definitive series – C.C.12 Oct 22 '19 at 17:01
2

Let $\ v=c-u,$ then $$I=\int\limits_{c-b}^{c-a}\dfrac{\sin(c-v)\,\mathrm dv}{\sqrt[3]{(v-\sin(c-v))^2}} = \int\limits_{c-b}^{c-a}\dfrac{\sin(c-v)v^{\mathbf-{\large 2\over3}\,\mathrm dv}}{\left(1-\dfrac{\sin(c-v)}v\right)^{\large 2\over3}},$$ $$I = \int\limits_{c-b}^{c-a} S(c,v)\,\mathrm dv,\tag1$$ where $$S(c,v) = \sum\limits_{k=0}^\infty(-1)^k\dbinom{-2/3}{k}\sin^{k+1}(c-v)\,v^{^{\Large -\frac{3k+2}3}},\tag2$$ $$\dbinom{-2/3}{k} = \dfrac1{k!}\dfrac{\Gamma\left(\dfrac13\right)}{\Gamma\left(\dfrac13-k\right)} = {(-1)}^k\cdot1\cdot\dfrac23\cdot\dfrac56\cdot\dfrac89\dots\dfrac{3k-1}{3k},\tag3$$

Binom -2/3

$$2^{2m-1}\sin^{2m}x = \dfrac12\binom{2m}m + \sum\limits_{j=1}^{m} (-1)^j\dbinom{2m}{m-j}\cos 2jx\tag4$$ (see also Wolfram Alpha test), $$4^m\sin^{2m+1}x = \sum\limits_{j=0}^{m}(-1)^j\dbinom{2m+1}{m-j}\sin (2j+1)x \tag5$$ (see also Wolfram Alpha test).

Therefore, $$S(c,v) = S_0(c,v)+S_1(c,v)+S_2(c,v),$$ where \begin{align} &S_0(c,v) = \sum\limits_{m=1}^\infty\dbinom{-2/3}{2m}4^{-m}\binom{2m}m\,v^{^{\Large -\frac{6m+2}3}},\\ &S_1(c,v) = \sum\limits_{m=1}^\infty\dbinom{-2/3}{2m}2^{-2m+1}\,v^{^{\Large -\frac{6m+2}3}}\sum\limits_{j=1}^{m} (-1)^j\dbinom{2m}{m-j}\cos 2j(c-v)\\ &S_2(c,v) = - \sum\limits_{m=0}^\infty\dbinom{-2/3}{2m+1}4^{-m}\,v^{^{\Large -\frac{6m+5}3}}\sum\limits_{j=0}^{m}(-1)^j \dbinom{2m+1}{m-j}\sin (2j+1)(c-v),\\ \end{align}

wherein $$S_0(c,v) = \dfrac{_1F_2\left(\dfrac13,\dfrac56,1,\dfrac1{v^2}\right)-1}{v^{^{\Large -\frac{2}3}}},\tag6$$ (see also Wolfram Alpha calculations), $$I_0(c,v) = \int S_0(c,v) dv = 3\sqrt[3]v \left(_2F_1\left(-\dfrac16,\dfrac13, 1, \dfrac1{v^2}\right) - 1\right) + \mathrm{constant}\tag7$$ (see also Wolfram Alpha integration).

Applying the exponent integral, one can get formulas for the even factors

$$\begin{align} &f_{m,j}(c,v) = \int \cos(2j(c-v))\, v^{\large- 2m-^2/_3}\,\mathrm dv = -\dfrac12 v^{\large - 2m+^1/_3}\\[4pt] & \times\left(e^{2ijc} \operatorname E^\,_{\large \,2m+^2/_3}(2ijv) + e^{-2ijc}\operatorname E^\,_{\large \,2m+^2/_3}(-2ijv)\right) +\mathrm{constant} \end{align}\tag8$$

and for the odd factors

$$\hspace{-30mu}\begin{align} &g_{m,j}(c,v) = \int \sin((2j+1)(c-v))\, v^{\large -2m - ^5/_3}\,\mathrm dv = \dfrac i2 v^{\large -2m-^2/_3}\\[4pt] &\times\Bigg(e^{i(2j+1)c}\operatorname E^\,_{\large \,2m+^5/_3}(i(2j+1)v) - e^{-i(2j+1)c} \operatorname E^\,_{\large \,2m+^5/_3}(-i(2j+1)v)\Big) +\mathrm{constant}. \end{align}\tag9$$

Therefore, the solution is

$$ \begin{aligned} &I= \left(I_0(c,v) + \sum\limits_{m=1}^\infty\dbinom{-2/3}{2m}2^{-2m+1}\,\sum\limits_{j=1}^{m} (-1)^j\dbinom{2m}{m-j}f_{m,j}(c,v)\right. \\[4pt] &\left. - \sum\limits_{m=0}^\infty\dbinom{-2/3}{2m+1}4^{-m}\,v^{^{\Large -\frac{6m+5}3}}\sum\limits_{j=0}^{m}(-1)^j \dbinom{2m+1}{m-j}g_{m,j}(c,v)\right)_{c-b}^{c-a}, \end{aligned}\tag{10} $$ where all used functions are defined by formulas $(7)-(9).$

On the other hand, the identity

$$E_\nu(z) = \dfrac{z^n}{(1-\nu)_n}E_{\nu-n}(z) - e^{-z}\sum\limits_{k=0}^{n-1}\dfrac{z^k}{(1-\nu)_{k+1}}$$

allows to attack obtained result. However, all my attempts were unsuccessful.

Yuri Negometyanov
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