Using Integral Substitution to Solve a Challenging Integration Problem

[chemphys1]

Homework Statement

sorry if question is unclear can't draw the integal sign out

Show that
Integral infinity-0 dz/((e^2z) - 1)^1/2 = integral 1- 0 dx/(1-x^2)^1/2 = pi/2

The Attempt at a Solution

I can get from the second integral to pi/2, as the second integral is sin^1(1) = pi/2

However, I do not understand how to go between these two integrals

infinity-0 dz/((e^2z) - 1)^1/2 = 1- 0 dx/(1-x^2)^1/2

I tried substituing things like x = z, but doesn't work. Can't see how you would change the limits from infinity to 0, to 1 to 0?

Maths is not my strong point so this could well be quite simple

any help much appreciated

 

[Curious3141]

Hint: the denominator can be expressed as $\displaystyle (e^z + 1)^{\frac{1}{2}}.(e^z - 1)^{\frac{1}{2}} = e^z.(1 + e^{-z})^{\frac{1}{2}}.(1 - e^{-z})^{\frac{1}{2}}$

Does that help?

 

[chemphys1]

Hint: the denominator can be expressed as $\displaystyle (e^z + 1)^{\frac{1}{2}}.(e^z - 1)^{\frac{1}{2}} = e^z.(1 + e^{-z})^{\frac{1}{2}}.(1 - e^{-z})^{\frac{1}{2}}$

Does that help?

Thank you for the help!

I've subsituted x = e^-z

so dz = dx/-e^-z

integral becomes

1/e^z(1+x)^1/2 (1-x^1/2) * dx/-e^-z

e^z*-e^-z = 1

so 1/(1+x)^1/2 (1-x^1/2) dx = 1/(1-x^2)^1/2

and then e^-z = x e^-infinity = 1 hence new limits 1 to 0

I think that works?

 

[Curious3141]

Thank you for the help!

I've subsituted x = e^-z

so dz = dx/-e^-z

integral becomes

1/e^z(1+x)^1/2 (1-x^1/2) * dx/-e^-z

e^z*-e^-z = 1

so 1/(1+x)^1/2 (1-x^1/2) dx = 1/(1-x^2)^1/2

and then e^-z = x e^-infinity = 1 hence new limits 1 to 0

I think that works?

Very difficult to read your post without LaTex.

But this: "e^z*-e^-z = 1" is an error, because the result should be negative one.

And that negative sign is important when you transform the bounds.