Landen transformation

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Problem 2.24.30 from Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. "Интегралы. Ряды." [ККЧ+03]More problems in Subsection 2.24.2 "Problems"

Prove that if F(k)\mathbf{F}(k) is the complete elliptic integral of the 1st kind (see problems 230 and 231§6231 \S 6), then

F(k)=π2∏n=1∞(1+kn),kn=1−1−kn−121+1−kn−12k0=k,∣k∣<1\mathbf{F}(k) = \frac{\pi }{2} \prod_{n = 1}^{\infty }\left(1+k_{n}\right), \quad k_{n} = \frac{1-\sqrt{1-k_{n-1}^{2}}}{1+\sqrt{1-k_{n-1}^{2}}} \quad k_{0} = k, \quad \left|k\right| < 1.

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Problem 2.6.230 from Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. "Интегралы. Ряды." [ККЧ+03]More problems in Subsection 2.6.2 "Problems"

The integral K(k)=∫0π/2dφ1−k2sin⁡2φ,∣k∣<1\mathbf{K}(k) = \int_{0}^{\pi / 2} \frac{d \varphi }{\sqrt{1-k^{2} \sin^{2} \varphi }},\left|k\right| < 1, is called the complete elliptic integral of the first kind in Legendre's form.

Let k1=1−1−k21+1−k2k_{1} = \frac{1-\sqrt{1-k^{2}}}{1+\sqrt{1-k^{2}}}. Prove that

K(k)=(1+k1)K(k1). \mathbf{K}(k) = \left(1+k_{1}\right) \mathbf{K}\left(k_{1}\right).
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Problem 2.6.231 from Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. "Интегралы. Ряды." [ККЧ+03]More problems in Subsection 2.6.2 "Problems"

Prove that if K(k)\mathbf{K}(k) is the complete elliptic integral of the first kind (see the previous problem), ∣k∣<1,k0=k\left|k\right| < 1, k_{0} = k, then:

kn=1−1−kn−121+1−kn−12,n=1,2,… k_{n} = \frac{1-\sqrt{1-k_{n-1}^{2}}}{1+\sqrt{1-k_{n-1}^{2}}}, \quad n = 1,2, \ldots
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(1)

lim⁡n→∞K(kn)=π2\lim_{n \rightarrow \infty } \mathbf{K}\left(k_{n}\right) = \frac{\pi }{2}

(2)

K(k)=π2lim⁡n→∞(1+k1)(1+k2)…(1+kn)\mathbf{K}(k) = \frac{\pi }{2} \lim_{n \rightarrow \infty }\left(1+k_{1}\right)\left(1+k_{2}\right) \ldots \left(1+k_{n}\right).

References

[ККЧ+03] Кудрявцев Л.Д., Кутасов А.Д., Чехлов В.И., Шабунин М.И. Интегралы. Ряды. 2-е изд., ФИЗМАТЛИТ, 2003.