MHB Definite integral on elliptic integral where modulus is function of variable

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The discussion focuses on proving the integral equation involving a definite integral of a function that includes the complete elliptic integral of the first kind. The integral in question is expressed as $\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}d\theta=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}}$, where $k(\theta)$ is defined in terms of the parameters R, h, and Z. Participants share their attempts to solve the integral, noting that the complexity increases with various methods tried. One participant reformulates the identity to a new expression involving $\sqrt{\sin \theta \cos \theta}$ and suggests that multiplying both sides by $2\sqrt{Rh}$ simplifies the problem. The conversation emphasizes the challenges faced in deriving the proof and invites further input on potential strategies.
bshoor
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How to prove:

$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}} $

where \[ k(\theta)=\sqrt\frac{4Rh \tan \theta}{Z^2+(R+h \tan \theta)^2}\]

and $ K[k(\theta)] $ is the complete elliptic integral of the first kind, defined by

\[ K[k(\theta)]= \int_0^{\frac{\pi }{2}}\frac{\,d\phi}{\sqrt{1-k^2(\theta)\sin^2 \phi}}\]

and h, R and Z $ \gt 0 $
 
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Can you show us what you have tried and where you are stuck? This will give our helpers a better idea how to provide help without perhaps offering suggestions that you may already be trying.
 
bshoor said:
How to prove:

$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}} $

where \[ k(\theta)=\sqrt\frac{4Rh \tan \theta}{Z^2+(R+h \tan \theta)^2}\]

and $ K[k(\theta)] $ is the complete elliptic integral of the first kind, defined by

\[ K[k(\theta)]= \int_0^{\frac{\pi }{2}}\frac{\,d\phi}{\sqrt{1-k^2(\theta)\sin^2 \phi}}\]

and h, R and Z $ \gt 0 $
Please make correction of the post:
$\int_{0}^{\frac{\pi }{2}} {\frac{\sin \theta}{\sqrt{Z^2+(R+h \tan \theta)^2}} K[k(\theta)]}d\theta=\frac{\pi }{2\sqrt{R^2 + (h+Z)^2}} $
 
MarkFL said:
Can you show us what you have tried and where you are stuck? This will give our helpers a better idea how to provide help without perhaps offering suggestions that you may already be trying.

I have tried in different ways. But the integral becomes more and more complicated. Anyway the identity can be modified to :

$\int_{0}^{\frac{\pi }{2}}{\sqrt {\sin \theta cos\theta}k(\theta) K[k(\theta)]}d\theta=\pi \sqrt{ \frac{Rh}{R^2 + (h+Z)^2}} $

by multiplying both side by $ 2 \sqrt{Rh} $
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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