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Evaluate $$ \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left[\frac{1+2r^2}{1-r^2\cos\left(2\theta\right)}\right]^{1/3}{\rm d}\theta $$ Question- Can I take the limit inside the integral?

My try-

$$I= \lim_{r \to 1^{-}} \int_{-\pi}^{\pi} \left(\frac{1+2r^2}{1-r^2\cos2\theta} \right)^{1/3}\, d\theta $$ $$ I= \int_{-\pi}^{\pi} \lim_{r \to 1^{-}} (\frac{1+2r^2}{1-r^2\cos2\theta})^{1/3}d\theta $$ $$ I=3^{1/3} \int_{-\pi}^{\pi} \frac{1}{(1-\cos2\theta)^{1/3}}d\theta $$ $$ I= 3^{1/3}2\int_{0}^{\pi} \frac{1}{(1-\cos2\theta)^{1/3}} d\theta $$ $$ I= 3^{1/3}4\int_{0}^{\pi/2} \frac{1}{(1-\cos2\theta)^{1/3}} d\theta $$ $$ I= (3/2)^{1/3}4\int_0^{\pi/2} \sin^{-2/3}\theta \ d\theta $$ $$I= 4(3/2)^{1/3} \frac{\Gamma(1/6)\Gamma(1/2) }{\Gamma(2/3)}. $$

Michael Hardy
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  • @Sid How to evaluate the integral? It is not easy for me –  Mar 01 '21 at 17:25
  • pardon me. i didnt see it was raised to the power 1/3 – Sid Mar 01 '21 at 17:29
  • @Sid Can't we interchange the limit and integral? –  Mar 01 '21 at 17:32
  • not sure... whats the source of this problem? is it some exam or? – Sid Mar 01 '21 at 17:34
  • @Sid It is from an exam. –  Mar 01 '21 at 17:37
  • Similar Question https://math.stackexchange.com/questions/253696/can-a-limit-of-an-integral-be-moved-inside-the-integral#:~:text=Taking%20the%20limit%20inside%20the,case%20of%20Dominated%20convergence%20theorem. – user6262 Mar 01 '21 at 18:13
  • @user1055 But here we have r tending to a finite limit . Also in the link given by you there are sequence of functions in the integrand which is not the case here. –  Mar 01 '21 at 18:16
  • I was actually taking about DCT and MCT not the integrand. – user6262 Mar 01 '21 at 18:23
  • Since the integrand is uniformly convergent we can take limit inside integral. – user6262 Mar 01 '21 at 18:34
  • @user 1055 Thanks. How is the integrand uniformly convergent? –  Mar 01 '21 at 18:50
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    the integrand is not uniformly convergent as the limit for $r=1$ has singularities at $0, \pm \pi$ (uniform convergence implies uniform boundness); however easy estimates show that all those points the $r$ integrand is bounded (uniformly in $r$) by a negative $2/3$ power (which is integrable) so by DCT we can switch limit and integral (eg near zero it is bounded by $c\theta^{-2/3}$, near $\pi$ by $c|(\theta-\pi)|^{-2/3}$ etc – Conrad Mar 01 '21 at 19:14
  • @Andrew John: A very interesting question: The integral for any r can be expressed by a sum of two hypergeometric Functions 2F1. Then the limit for r = 1 delivers just the half of your answer, means 16.6807 – stocha Mar 02 '21 at 09:41
  • @stocha Can you please elaborate –  Mar 02 '21 at 13:08
  • Now the idea is to first calculate the integral for any r and then take the limit. I was encountered with this integral by the study of elliptic integrals, I just did the integration by mathematica, because I did not find the paper yet, and omit to post the solution. The variable substiutions are hard to find, but in the end one gets an Appell F1 and reduce it to the sum of two 2F1 functions. – stocha Mar 02 '21 at 21:40
  • @SHEKHAR SUMAN e.g.https://www.researchgate.net/publication/228527641_On_a_Class_of_Generalized_Elliptic-type_Integrals – stocha Mar 02 '21 at 23:24
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    You have an erroneous factor of $2$ because $\int_0^{\pi/2}\sin^{2a-1}\theta\cos^{2b-1}\theta d\theta=\tfrac12\operatorname{B}(a,,b)$. – J.G. Mar 03 '21 at 19:14

1 Answers1

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This is not an unswer, just a hint and too long for a comment: The integral above may be transformed with $\theta = \frac{\arccos(y)}{2}$

$$2\, \left(1+2\, r^2\right)^{1/3} \int_{-1}^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dy$$ Next we can split the integral in two parts with $p=-y$ in the first one:

$$2\,\left(1+2\, r^2\right)^{1/3} \left(\int_0^1 \left(1+r^2 p\right)^{-1/3} \left(1-p^2\right)^{-\frac{1}{2}} \, dp + \int_0^1 \left(1-r^2 y\right)^{-1/3} \left(1-y^2\right)^{-\frac{1}{2}} \, dp \right) $$

and evaluate them in form of hypergeometric functions. For this step e.g. we can express the integrand in form of MeijerG - Function. The first integral will be transformed with $q = p^2$ apply the formular Mathematical Functions Site and simplify:

$$I_1 = \frac{\left(1+2\, r^2\right)^{1/3}}{\gamma\left[\frac{1}{3}\right]} \int_0^1 (1-q)^{-\frac{1}{2}} q^{-\frac{1}{2}} \,G_{1,1}^{1,1}\left( \sqrt{q}\, r^2\left\vert \begin{array}{c} \frac{2}{3}\\ 0% \end{array}% \right. \right)\, \, dq$$ $$ = \left(1+2 \,r^2\right)^{1/3} \left(\pi \,_{2}F_{1}\left[\frac{1}{6},\frac{2}{3},1,r^4\right]-\frac{2}{3} r^2 \,_{3}F_{2}\left[\left\{\frac{2}{3},1,\frac{7}{6}\right\},\ \left\{\frac{3}{2},\frac{3}{2}\right\},r^4\right]\right)$$

The second integral $I_2$ is performed in the same way. The limit $r \to 1$ Mathematical Functions Site and Mathematical Functions Site exists e.g. for $I_1$ $$\lim_{r \to 1}\,I_1=\frac{\frac{3 \pi\, \gamma\ \left[\frac{1}{6}\right]}{\gamma\left[\frac{1}{3}\right]\, \gamma\left[\frac{5}{6}\right]}-2\,_{3}F_{2}\left[\left\{\frac{2}{3},1,\frac{7}{6}\right\},\left\{\frac{3}{2},\frac{3}{2}\right\},1\right]}{3^{2/3}}$$ and for $I_2$ in the same way. The sum of $I = I_1 + I_2$ delivers just the half of your answer:

$$I= \frac{2^{1/3}\, 3^{5/6}\, \gamma\left[\frac{1}{3}\right]^2}{\gamma\left[\frac{2}{3}\right]} $$.

Therefore, inspite of a calculation error of factor 2, @user1055 is right, the limit may be also calculated inside the integral.

stocha
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