elliptic integral

kexue

When replacing x with 1/kx then

[tex]
\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 }}
[/tex]

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.

Attached Files

mapping.pdf (149.6 KB, 12 views)

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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