Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I How to evaluate this integral?

  1. Aug 22, 2016 #1
    Hello all,

    I need to evaluate this integral

    [tex]\int_0^{\infty}(1+\alpha_2)^{-2}\left(\alpha_2+b\right)^{-1}\,d\alpha_2[/tex]

    I found the following integral in the table of integrals (eq 3.197.1, 7th edition)

    [tex]\int_0^{\infty}x^{v-1}(\beta+x)^{-\mu}(x+\gamma)^{-\rho}\,dx=\beta^{-\mu}\gamma^{\mu-\rho}B(v,\mu-v+\rho)_2F_1(\mu,\,v;\,\mu+\rho;1-\frac{\gamma}{\beta})[/tex]

    The integral and the conditions are in the image attached. Comparing the two integrals I set ##v=1,\,\beta=1,\,\gamma=b##, which implies that ##\mu=2## and ##\rho=1##. I satisfy the last two conditions, but I'm not sure about the first two. What does it mean arg x if x is a constant?

    ##b## varies in a loop, and when I evaluate the integral I get results. However, the final result, in which the above integral is a part, isn't right. I'm not sure where the error is, and I wanted to make sure it isn't in using the integral.

    I appreciate any help. Thanks
     

    Attached Files:

  2. jcsd
  3. Aug 22, 2016 #2

    Mark44

    Staff: Mentor

    It doesn't say arg(x) in the integral you found -- it says arg(β) and arg(γ). This implies that these parameters are complex. In the integral x is real.

    The integral you found should work, but I don't know what the B and F1 functions are. The integral could be evaluated using much simpler means, though, by using partial fractions decomposition to split the integrand into three parts.

    The decomposition would look like this:
    $$\frac{1}{(a + 1)^2} \frac{1}{a + b} = \frac{A}{a + 1} + \frac{B}{(a + 1)^2} + \frac{C}{a + b} $$
    After you determine the constants A, B, and C, integrate each term and then evaluate your three improper integrals.
     
  4. Aug 22, 2016 #3

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    The equation is valid for some complex numbers as well, the condition on the complex argument are satisfied for positive real values (and also for all complex values with a positive imaginary component).

    It is not necessary to use such a powerful formula, for integer exponents you can use partial fraction decomposition.
     
  5. Aug 22, 2016 #4
    Thanks, but why ##\frac{A}{a + 1} + \frac{B}{(a + 1)^2} + \frac{C}{a + b} ## and not ##\frac{A}{(a + 1)^2} + \frac{B}{a + b} ##? I did the latter, and I got the following:

    [tex]\frac{1}{(x+1)^2(x+b)}=\frac{1}{b-1}\left[\frac{1}{(x+1)^2}-\frac{1}{x+b}\right][/tex]

    Then I used the integral formula (attached):

    [tex]\int_0^{\infty}\frac{x^{\mu-1}}{(1+\beta\,x)^v}\,dx=\beta^{-\mu} B(\mu,\,v-\mu)[/tex],

    where ##B(.,.)## is the beta function. I got the following:

    [tex]\int_0^{\infty}\frac{1}{(x+1)^2}\,dx=B(1,1)[/tex]

    and

    [tex]\int_0^{\infty}\frac{1}{(x+b)}\,dx=B(1,0)[/tex]

    But evaluating ##B(1,1)-B(1,0)## in Mathematica gives me ComplexInfinity!! Why?
     

    Attached Files:

  6. Aug 22, 2016 #5

    Mark44

    Staff: Mentor

    Because that's how you do it when you have repeated linear factors (i.e., (a + 1)2). If you break it up into only two terms, you don't get the correct decomposition. You need to break the original integrand into three terms, not two.
     
  7. Aug 22, 2016 #6
    Actually, I did it as you said, and got A=0, which led to the same result as I did it!!
     
  8. Aug 22, 2016 #7

    Mark44

    Staff: Mentor

    Then you made a mistake -- A is not 0. I'm pretty confident in this, as I checked my work.

    If you show what you did, we can figure out where you went wrong.
     
    Last edited: Aug 22, 2016
  9. Aug 22, 2016 #8
    OK. So first we have

    [tex]
    \frac{1}{(x + 1)^2} \frac{1}{x + b} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2} + \frac{C}{x + b}=\frac{A(x+1)(x+b)+B(x+b)+C(x+1)^2}{(x+1)^2(x+b)}
    [/tex]

    This implies that

    $$A(x+1)(x+b)+B(x+b)+C(x+1)^2=1$$

    For ##x=-1## we get ##B=\frac{1}{b-1}##

    For ##x=-b## we get ##C=-\frac{1}{b-1}##

    For ##x=0## we get ##Ab+Bb+C=Ab+\frac{b}{b-1}-\frac{1}{b-1}=Ab+1=1## which implies that ##A=0##.

    Where did I err?
     
  10. Aug 22, 2016 #9

    Mark44

    Staff: Mentor

    In your calculation for C. I get C = ##-\frac{1}{(b - 1)^2}##
     
  11. Aug 22, 2016 #10
    I think in this case ##C=\frac{1}{(b-1)^2}##, isn't it? But you are right, I missed the square, and don't know how I missed it THREE times!! Anyway, thanks
     
  12. Aug 22, 2016 #11

    Mark44

    Staff: Mentor

    Correct. I misread my work.
    One little mistake will throw everything off. That's why I checked my work. After substituting the values for A, B, and C, I ended up with ##\frac{(b - 1)^2}{(b - 1)^2} \equiv 1## (except if b = 1).
     
  13. Aug 23, 2016 #12
    The integral I attached in post #4 works only for ##(x+1)^2##. How to evaluate the other integrals? I found this indefinite integral

    $$\int\frac{1}{ax+b}\,dx=\frac{1}{a}\ln|ax+b|$$

    Can I use this?
     
  14. Aug 23, 2016 #13

    Mark44

    Staff: Mentor

    Yes. For both integrals a = 1.
     
  15. Aug 23, 2016 #14
    Right, but the integrals in my case are definite. Do I need to evaluate the result as ##\left.
    \frac{1}{a}\ln|ax+b|\right|_0^{\infty}
    ##. This will result in ##+\infty##!!
     
  16. Aug 23, 2016 #15

    Mark44

    Staff: Mentor

    All three are improper integrals, so you can't simply substitute ##\infty## into the antiderivatives. You'll need to use limits. Any calculus textbook should have a section on dealing with improper integrals.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: How to evaluate this integral?
Loading...