1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Integral Substitution Question

  1. Dec 30, 2015 #1
    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. jcsd
  3. Dec 30, 2015 #2


    User Avatar
    Science Advisor
    Homework Helper

    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.
  4. Dec 30, 2015 #3
    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!!
  5. Dec 30, 2015 #4


    User Avatar
    Science Advisor
    Homework Helper

    Glad you found it. Don't feel irked, that's the eternal joy of mathematics.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted