Integral Substitution Question

[Bill333]

Homework Statement

Show by appropiate substitutions that ∫ (e^2z-1)^-0.5 dz from 0 to infinity is equivalent to ∫(1-x²)^-0.5 dx from 0 to 1. Thus, show that the answer is π/2.

Homework Equations

The Attempt at a Solution

Where to begin! I tried the substitution e^2z= 2-x², but this then transforms the integral into ∫(2-x²)/(x(1-x²)^0.5) dx, and the limits don't work as it would then require the root of a negative number.
I then tried the substitution e^z = x, but that gives you ∫1/(x(x²-1)^0.5) dx, the limits being infinity and 1.
I carried on a bunch of different substitutions involving the e^z and a simple x polynomial, which I won't repeat due to them being abysmal failures!
In doing the problem I managed to skip the middle step entirely, and got the value of π/2 I believe by re-writing the initial equation as -∫(e^2z-1)/(e^2z-1)^0.5 dz +0.5∫2e^2z/(e^2z-1)^0.5 dz, both still from 0 to infinity. The first of the pair became -∫(e^2z-1)^0.5 dz between 0 and infinity, which solves to become -[(e^2z-1)^0.5)-sec^-1(e^z)] between infinity and 0. The second part I performed the substitution x=e^2z-1, with dx = 2e^2z dz. Thus this was transformed into (0.5)∫1/(u^0.5) du between 0 and infinity. This integrated to give [u^0.5] between 0 and infinity, which was then equal to (e^2z-1)^0.5. The sum of these two parts gives [sec^-1(e^z)] between 0 and infinity which is equal to cos^-1(1/e^z) between 0 and infinity, which gives π/2.
Unfortunately the question specifies the middle step has to be achieved :(.
Onto my current ideas, I thought using a substitution along the lines of z= (2/π)tan^-1(x), which transforms the limits into 1 and 0, and transforms the integral into ∫1/((1-x²)(e^{4(tan-1(x)/π})^0.5) dx. This has the 1-x² part in it, but it is not square rooted and I feel this is as far as I could go.

Any help concerning which substitutions would be appropriate would be very much appreciated :)!

 

[Samy_A]

You are overthinking it.

Look at how the integration limits have to change. $e^z=x$ doesn't work (as you correctly noticed), but something similar could.

 

[Bill333]

Ah, is the substitution instead e^z = 1/x, thus -(x)^-2 dx = e^zdz, thus -1/x dx = dz.
The integral is now ∫(-1/(x²-1)^-0.5*x) dx from 1 to 0, which can be re-written as ∫x/(x²-1)^0.5*x) from 0 to 1, which then cancels to give the form as required.

Thank you very much for your help, I feel irked now that I didn't see it!

 

[Samy_A]

Ah, is the substitution instead e^z = 1/x, thus -(x)^-2 dx = e^zdz, thus -1/x dx = dz.
The integral is now ∫(-1/(x²-1)^-0.5*x) dx from 1 to 0, which can be re-written as ∫x/(x²-1)^0.5*x) from 0 to 1, which then cancels to give the form as required.

Thank you very much for your help, I feel irked now that I didn't see it!

Glad you found it. Don't feel irked, that's the eternal joy of mathematics.