# Integral Substitution Question

1. Dec 30, 2015

### Bill333

1. The problem statement, all variables and given/known data
Show by appropiate substitutions that ∫ (e2z-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.

2. Relevant equations

3. The attempt at a solution
Where to begin! I tried the substitution e2z= 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 ez = 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 ez 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 -∫(e2z-1)/(e2z-1)0.5 dz +0.5∫2e2z/(e2z-1)^0.5 dz, both still from 0 to infinity. The first of the pair became -∫(e2z-1)0.5 dz between 0 and infinity, which solves to become -[(e2z-1)0.5)-sec-1(ez)] between infinity and 0. The second part I performed the substitution x=e2z-1, with dx = 2e2z dz. Thus this was transformed into (0.5)∫1/(u0.5) du between 0 and infinity. This integrated to give [u0.5] between 0 and infinity, which was then equal to (e2z-1)0.5. The sum of these two parts gives [sec-1(ez)] between 0 and infinity which is equal to cos-1(1/ez) 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)(e4(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 :)!!

2. Dec 30, 2015

### 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.

3. Dec 30, 2015

### Bill333

Ah, is the substitution instead ez = 1/x, thus -(x)-2 dx = ezdz, 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!!

4. Dec 30, 2015

### Samy_A

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