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I Is this integral integrable?

  1. Apr 14, 2016 #1
    After following the logical steps to derive something, I reached to the following integral:

    [tex]\int_0^{\infty}\frac{1}{x^2}\exp\left(\frac{A}{x}-x\right)E_1\left(B+\frac{A}{x}\right)\,dx[/tex]

    where ##E_1(.)## is the exponential integral function, and ##A## and ##B## are non-zero positive constants. I tried to find this integral in the table of integrals, but couldn't find it. Next best thing to do is to evaluate this integral numerically, I guess. So, I resorted to Mathematica for that purpose, and used the Nintegrate command, but I got a message that the integrand has evaluated to overflow, indeterminate, or infinity. It suggested to increase the number of recursive refinements, but it didn't work. Is there something fundamentally wrong in the integral, or it's just a technical issue, and how to overcome it? I have to mention that when I replaced the lower bound of the integral by a very small value that is greater than 0, the integral was calculated perfectly. Is this the problem maybe?

    Thanks
     
  2. jcsd
  3. Apr 14, 2016 #2
    Many symbolic algebra systems have this issue with their numerical functions. When there are [itex]\frac{1}{x}[/itex] terms and such inside the expression, even if the limit of the entire expression is finite, the numerical algorithm fails early when evaluating the expression from inside out. There are slightly kludgy looking solutions that usually work. I personally like to add a while-type conditional to the expression which forces it to evaluate to the proper limit at the points in question.
     
  4. Apr 14, 2016 #3
    How do you add "a while-type conditional to the expression"?
     
  5. Apr 14, 2016 #4
    It depends on the particular CAS you're using. It's the same as an if-then-else structure if that helps.

    For example, in some CAS languages, while(x=0,0,exp(-1/x^2)) will evaluate to 0 when x is 0 and to exp(-1/x^2) everywhere else. Look up how to do piecewise functions in your language, it's the same thing. In this case, one of the pieces of the domain is a single point instead of an interval, is all.
     
  6. Apr 14, 2016 #5

    George Jones

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    In order to evaluate the integral numerically, what values of A and B do you use?
     
  7. Apr 14, 2016 #6
    I need to evaluate the integral at different values of A and B, but let's say A=50 and B=5 as one of the cases.
     
  8. Apr 18, 2016 #7
    Any hint?
     
  9. Apr 25, 2016 #8
    The integral diverges at ##0##, the integrand behaves near ##0## as ##1/(Ax)##, which is not integrable.
     
  10. Apr 25, 2016 #9
    Can we approximate it such that it becomes integrable?
     
  11. Apr 26, 2016 #10
    What do you mean? Your function is positive, non integrable, so integral is ##+\infty##, there is no other reasonable interpretation. For non-integrable functions than change sign, there are some possibilities to define a finite value of the integral, for example, the principal value integral, but for positive functions there is no alternative interpretation, the integral is ##+\infty##.

    As for the "approximation" you can change your integrand to get a finite number, but this will not have any relation to the original integral.
     
  12. Apr 26, 2016 #11
    OK. Now if I followed the mathematical derivations for something correctly, such that this something is between 0.5 and 1 for sure (I can get this using Monte-Carlo simulations), is it possible to reach to such an integral that is not integrable?
     
  13. Apr 26, 2016 #12
    You probably made mistake in your calculations.
     
  14. Apr 26, 2016 #13
    OK. Could I walk you through the derivation. Would you patient with me?
     
  15. Apr 26, 2016 #14
    OK, I'll try to follow.
     
  16. Apr 26, 2016 #15
    OK. perfect. Now I have something I need to finish. Tomorrow, I will start the derivation from the beginning to see where I erred. Thanks
     
  17. Apr 28, 2016 #16
    Hi. Sorry, I was busy yesterday. Basically, I have this expession

    [tex]\varepsilon(\alpha_1,\,\alpha_2,\,\alpha_3)=\exp\left(-\frac{\frac{\alpha_1}{\alpha_2}A}{\alpha_3 B+1}\right)[/tex]

    where ##\alpha_i## for i=1, 2, 3 are independent and identically distributed exponential random variables with mean 1, and A and B are non-zero positive constants. I need to find the average value of ##\varepsilon(\alpha_1,\,\alpha_2,\,\alpha_3)##. To do so, I let ##X=\frac{\frac{\alpha_1}{\alpha_2}A}{\alpha_3 B+1}## and then found the cumulative distribution function (CDF) of X denoted by ##F_X(x)##, and then found its derivative with respect to ##x## to find the probability density function (PDF), denoted by ##f_X(x)##. After finding the PDF, the average value of ##\varepsilon(\alpha_1,\,\alpha_2,\,\alpha_3)## can be found as

    [tex]\varepsilon=\int_0^{\infty}e^{-x}\,f_X(x)\,dx[/tex]

    I will stop here to see everything I did so far is correct, and to make sure you're following. I'll provide the details of ##F_X(x)## and ##f_X(x)## in the next post.
     
  18. Apr 28, 2016 #17
    The expectation should be $$\int_{-\infty}^\infty x f_X(x) dx,$$ see Wikipedia. Your quantity is positive, so ##f_X(x)=0## for ##x<0##, so you can integrate only from ##0## to ##\infty##, but ##e^{-x}## is wrong.
     
  19. Apr 28, 2016 #18
    But I need to find the expected value of ##\varepsilon## not ##X##!! Right?
     
  20. Apr 28, 2016 #19
    Yes, you are right. Sorry, my bad here.
     
  21. Apr 28, 2016 #20
    OK, perfect. So first I need to find the CDF as

    [tex]F_X(x)=\text{Pr}\left[X\leq x\right]=\text{Pr}\left[\frac{\frac{\alpha_1}{\alpha_2}A}{\alpha_3 B+1}\leq x\right]=\int_{\alpha_2=0}^{\infty}\int_{\alpha_3=0}^{\infty}\text{Pr}\left[\alpha_1\leq\frac{x\alpha_2}{A}\left(\alpha_3 B+1\right)\right]f_{\alpha_2}(\alpha_2)f_{\alpha_3}(\alpha_3)\,d\alpha_2d\alpha_3=\int_{\alpha_2=0}^{\infty}\int_{\alpha_3=0}^{\infty}\left[1-\exp\left(-\frac{x\alpha_2}{A}\left(\alpha_3 B+1\right)\right)\right]e^{-\alpha_2}e^{-\alpha_3}d\alpha_2d\alpha_3=1-\frac{A}{xB}\int_{\alpha_3=0}^{\infty}\frac{e^{-\alpha_3}}{\alpha_3+\frac{x+1}{B}}\,d\alpha_3[/tex]

    From the table of integral we have

    [tex]\int_0^{\infty}\frac{e^{-\mu x}}{x+\beta}\,dx=-e^{\beta\mu}\text{Ei}(-\beta\mu)[/tex]

    where ##\text{Ei}## is the exponential integral function. But then I used the property that ##E_1(z)=-\text{Ei}(-z)##, which results in

    [tex]F_X(x)=1-\frac{A}{xB}\exp((x+1)/B)E_1((x+1)/B)[/tex].

    Is everything OK so far?
     
    Last edited: Apr 28, 2016
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