Help me prove $K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$ where $K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$

Question

Hi this problem is posed in Sean M. Stewart's How to Integrate It book pg.143

$$K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}}$$ where $ 0 \leq k < 1$

Based on the substitution:

$$ \tan{u} = \frac{\sin{2\theta}}{k+k\cos{2\theta}} =\frac{\tan{\theta}}{k} $$

Show that:

$$K(k) = \frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$$

$K(k) $ is some function defined in term of definite-integral called complete elliptic integral which is not easily computable, so attacking $K(k)$ gave no fruitful result.

Well then proceeding with substitution ( It is easy to lose track as it involved a lot of trigonometric manipulations )

I reduced the integral into this form : $$K(k) = k\int^{\pi/2}_0 \frac{\sec^2{\theta}d\theta}{\sqrt{k^2 + \tan^2{\theta}}\sqrt{k^2 + \tan^2{\theta} - k^2\tan^2{\theta}}}$$

Using substitution : $\tan{\theta} = p$

$$K(k) = k\int^{\infty}_0 \frac{dp}{\sqrt{k^2 +p^2}\sqrt{k^2 +p^2 - (kp)^2}}$$

To my naive eyes this looked manageable but WolframAlpha says otherwise.

Any hint or ideas on how to attack this problem?

I thought the trig manipulations to find a more suitable form but in the I end up transform $\sin {u}$ term into $\tan{u}$ to use the given substitution

EDIT: It has been suggested in commens that there is typo in the substituion(given in book) , it should be :

$$ \tan{u} = \frac{\sin{2\theta}}{k+\cos{2\theta}} $$

Answer 1

Here is my solution based on Paramanand Singh's hint:

First convert integral from [0,π/2] to [0,π] and then the interval [0,π] of u maps to [0,π/2] of θ

Using the definite integral property :

$\int^{2a}_0 f(x) dx = 2\int^a_0 f(x)dx$ if $f(2a-x) = f(x)$

We change the limits to : Upper Limit $ = \pi$ and lower limit $= 0$ which will allow us to use the substitution given in the book [ $$ ]

$$K(k) = \int^{\pi/2}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}} = \frac{1}{2}\int^{\pi}_0 \frac{du}{\sqrt{1-(k\sin{u})^2}} $$

As the substitution is in the form of $\tan u $ we will express the integrand in terms of $\tan u$

We arrive at :

$$\frac{1}{2}\int^{\pi}_0 \frac{\sec{u}du}{\sqrt{1 + \tan^2{u}-(k\tan{u})^2}} $$

Substituting : $\tan{u} = \frac{\sin{2\theta}}{k+\cos{2\theta}}$ [ $$ ]

$$\frac{1}{2}\int^{\pi}_0 \frac{(k+ \cos{2\theta})}{k\cos{2\theta} +1}\sec{u}du $$

Now,

$$\sec^2{u}du = \frac{2(k\cos{2\theta}+1)d\theta}{(k + cos{2\theta})} $$

Leading to a cleaner integral:

$$\int^{\pi/2}_0 \frac{d\theta}{(k+\cos{2\theta})\sec{u}}$$

As

$\sec u = \sqrt{1+ \tan^2{u}} = \frac{\sqrt{k^2 + 1 + 2k\cos{2\theta}}}{(k + \cos{2\theta})} = \frac{\sqrt{(k+1)^2 - 4k\sin^2{\theta}}}{(k + \cos{2\theta})} $

Integrand reduces to desired form:

$$ \int^{\pi/2}_0 \frac{d\theta}{\sqrt{(k+1)^2-(2\sqrt{k} \sin{\theta})^2}} $$

I think it is trivial to take from here to the desired form :)

[ $$ ] If you are coming here from the book, this problem passage had lot of minor typos, rely on this post as a reference.

For.eg

For this substitution it can be shown that the limits of integration change in the following manner. When u = $0$, $\theta$ = $0$, and when u = $\pi/2$, $\theta= \pi/2$

This is incorrect

Answer 2

I think that you are on the right track. But you have to be aware that the completed elliptic integral of this type can be defined by any of three integrals:

$$ K(k)=\int_0^{\pi/2}\frac{1}{\sqrt{1-k^2\sin^2\theta}}\ d\theta\\=\int_0^1\frac{1}{\sqrt{(1-t^2)(1-k^2t^2)}}\ dt\\ =\int_0^\infty \frac{1}{\sqrt{(1+t^2)[(1+(1-k^2)t^2)}]}\ dt $$

This is from K. Oldham, J. Myland, & J. Spanier, An Atlas of Functions, Springer . Chapter 61 covers the Complete Elliptic Integrals. Can you take it from here?

Answer 3

Using the Binomial Theorem: $$\sum_{n=0}^{\infty}\frac{(2n)!x^{n}}{2^{2n}n!^2}=\frac{1}{\sqrt{1-x}} \tag{1}$$ The expansion for $K(k)$ wich follows after replacing $x=k^2\sin^2{x}$ in $(1)$, integrating in $(0,\pi/2)$ and aplying Wallis' formula $\int_{0}^{\pi/2}\sin^{2n}{(x)}dx=\frac{\pi}{2}\frac{(2n)!}{2^{2n}n!^2}$: $$\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n)!^2k^{2n}}{2^{4n}n!^4}=\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}=K(k) \tag{2}$$ Then replacing in $(1)$ with $x=-ke^{ix}$ using $(2)$ and Parseval's Theorem: $$\frac{2}{\pi}K(k)=\sum_{n=0}^{\infty}\left(\frac{(2n)!(-k)^n}{2^{2n}n!^2}\right)^2=\frac{1}{\pi}\int_{-\pi}^{\pi}|{\frac{1}{\sqrt{1+ke^{ix}}}}|^2dx \\= \frac{1}{\pi}\int_{0}^{\pi}\frac{dx}{\sqrt{2k\cos{x}+k^2+1}}=\frac{2}{\pi}\int_{0}^{\pi/2}\frac{dx}{\sqrt{-4k\sin^2{x}+(k+1)^2}} \\=\frac{2}{\pi}\frac{1}{1+k}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1-\left(\frac{2\sqrt{k}}{1+k}\right)^2\sin^2{x}}}=\frac{2}{\pi}\frac{K(\frac{2\sqrt{k}}{1+k})}{1+k}.$$ Where we have used $\cos{2x}=1-2\sin^2{x}$ in the fifth step. And we are done. $$K(k)=\frac{1}{1+k}K(\frac{2\sqrt{k}}{1+k}).$$

Answer 4

Hints: Make clever substitute $$u=x+\arctan\left(\frac{(1-k)\tan x}{(1+k)}\right)$$ Then $$u\rightarrow\pi;\; x\rightarrow\frac{\pi}{2}$$ $$u\rightarrow0;\; x\rightarrow0$$