Solving Elliptic Integral: Replacing x with 1/kx

[kexue]

When replacing x with 1/kx then

$$\int_{1/k}^\infty {\left[ {\left( {x^2 - 1} \right)\left( {k^2 x^2 - 1} \right)} \right]} ^{ - 1/2} dx = \int\limits_0^1 {\left[ {\left( {\frac{1}{{k^2 x^2 }} - 1} \right)\left( {\frac{1}{{x^2 }} - 1} \right)} \right]} ^{ - 1/2} \frac{{dx}}{{kx^2 }}$$

I do not see how. Why ranges the LHS integral over infinity, whereas the RHS from 0 to 1?

Any help and hints very much appreciated.

thanks

 

[HallsofIvy]

As x goes from 1/k to infinity, kx goes from 1 to infinity, and 1/kx goes from 1 to 0. Make the change of variable u= 1/kx and then, since both u and x are "dummy variables", replace u with x.

 

[kexue]

thank you

 

[kexue]

From the same book, I have the following.

Arcsin (which is given its integral form) maps the upper half plane 1:1 onto the shaded strip |x|<pi/2, y>0.

Now the sentence I don't get. By reflection in the punctured plane (punctured at +1 and -1), it produces a full tiling of the target plane by congruent, nonoverlapping images of the upper and lower half-planes.

So by this reflection we get many strips, that then completely cover the target plane.

But how does that work?

thank you

 

[kexue]

I have attached a file.

What goes on with 'By reflection in the punctured plane (punctured at +1 and -1), it produces a full tiling of the target plane by congruent, nonoverlapping images of the upper and lower half-planes'.

I do not understand this.

 

[kexue]

Hellllooooooooo!

By the way, it is from the book 'Elliptic Curves' by McKean and Moll, p. 71. They call this example simple and a warm up.

Anybody out there that can give a comment?

 

[kexue]

don't be shy

 

[kexue]

What is it what you people here don't like about my question?

Please talk to me.

 

[kexue]

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