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My teacher said that calculate the following integral, and that the integral is convergent, I tried to calculate but failed, thanx in advanced for any help.

$$0<k<1$$ $$I=\displaystyle\int_0^1\dfrac{dx}{\sqrt{(1-x^2)(1-x^2k^2)}}=\text{?}$$

Attempt 1:

Integral's interval's boundaries are $1,0$ so I want to try $x=1-u$ substitution $$I \int_0^1\dfrac{dx}{\sqrt{(1-x^2)(1-x^2k^2)}}=\int_1^0 \frac{-du}{\sqrt{(1-(1-u)^2) (1-(1-u)^2k^2)}}=\text{?}$$

I could not get from this, denominator will get too complicated so we could not get benefit by this trick.

Attempt 2:

We can use odd and even function for symmetric intervals of integrals so the function inside the integral is even function and interval is $[0,1]$ that is if we extend the integral's interval to $[-1,1]$ and if we think $$\frac 1 {\sqrt{(1-x^2)(1-x^2k^2)}} = \frac{g(x)+g(-x)}{2}\tag{*},$$ we can rewrite the integral as follows:

$$I= \int_{-1}^1 g(x)\,dx$$ But I couldnot arrange the function $g$ from $(*)$

Other attemts:

I've tried to substitute trigonometric functions, $x=\sin y$, $x=\tan y$ and $x=uk$ and $x=u/k$

Michael Hardy
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Micheal Brain Hurts
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    This is the complete elliptic integral of first kind, no simple formula is known for general $k$. – pisco Jan 07 '18 at 17:18
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    You won't be able to evaluate this in closed form, but you should be able to show its value is finite. – Michael Hardy Jan 07 '18 at 17:25
  • In additional, I write both deminoters in series form and got last cauchy product form, but again I didnt get the bottom.

    How can you decide this can not be written in closed form, so fast, I can not still see now .

    Thank you for comments.

     – Micheal Brain Hurts Jan 07 '18 at 17:32
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    @RakiBalikMatematik - This is just a known integral, which is called the "incomplete elliptic integral of the first kind" and usually denoted $F(x,k)$, which in your case $x=1$. Elliptic integrals don't usually have closed forms, else we wouldn't need special names for their values. – Nathaniel Bubis Jan 07 '18 at 17:48

1 Answers1

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$I(k)$ is a complete elliptic integral of the first kind. It has some special values, and its numerical computation can be performed by exploiting the relation with the $\text{AGM}$ mean, defined in the following way.

Let $a,b>0$. The sequences $\{a_n\}_{n\geq 0},\{b_n\}_{n\geq 0}$ defined by $a_0=a, b_0=b$, $a_{n+1}=\frac{a_n+b_n}{2}$, $b_{n+1}=\sqrt{a_n b_n}$ converge (pretty fast) to a common limit, $\text{AGM}(a,b)$.

In our case we have

$$ I(k)=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-k^2\sin^2\theta}}=\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-k^2\cos^2\theta}}=\int_{0}^{+\infty}\frac{dt}{\sqrt{(1+t^2)(1-k^2+t^2)}} $$ and $$ I(k)=\color{red}{\frac{\pi}{2\cdot\text{AGM}(1,\sqrt{1-k^2})}}. $$ In terms of hypergeometric functions we have $$ \forall k\in(0,1),\qquad I(k)=\frac{\pi}{2}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{16^n}\,k^{2n} $$ and the asymptotic behaviour as $k\to 1^-$ is $-\frac{1}{2}\log\left(\frac{1-k}{8}\right)+o(1-k)$.

See also the section Elliptic Integrals and the AGM in my notes.

Jack D'Aurizio
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