Looking to solve an integral of the form $\int_1^\infty (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} \; dy$

1,013

Solution 1

Integrand behaves as $\frac{C}{y}$ as $y\rightarrow\infty$; hence the integral diverges.

Indeed, $$ (y-1)^{n-1} y^{-n} \sim y^{-1}$$ as $y\to\infty$, and $$ e^{(\alpha -\alpha n)\frac{(y-1) }{y}}\to e^{(\alpha -\alpha n) } $$

Solution 2

$$\begin{align}\int_1^\infty dy\, (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} &= e^{-\alpha (n-1)} \int_0^1 \frac{du}{u^2} \left (\frac1{u}-1 \right )^{n-1} u^n e^{\alpha (n-1) u} \\ &= e^{-\alpha (n-1)} \int_0^1 du (1-u)^{n-1} u^{-1} e^{\alpha (n-1) u} \end{align}$$

The integral diverges for all values of $n$.

Solution 3

\begin{align} \int_1^\infty (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} \; dy \\=\int_1^\infty \frac{(y-1)^{n-1}}{y^{n-1}y} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} \; dy \\=\int_1^\infty {(1-\frac{1}{y})^{n-1}}{\frac{1}{y}} e^{(\alpha -\alpha n){(1-\frac{1}{y}})} \; dy \end{align}

Let z = -ln(y) then dz = -dy/y. So $\frac{1}{y} = e^{z}$.

When z = 1, y = 1, and when z = $\infty$, y = 0.

\begin{align} \\=\int_0^1 {(1- e^z)^{n-1}}e^{(\alpha -\alpha n){(1-e^z})} \; dz \\=\int_0^1 {(1- e^z)^{n-1}}e^{(\alpha -\alpha n){(1-e^z})} \; dz \end{align}

Please check the above for any error during substitution.

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Updated on August 01, 2022

Comments

  • Nero
    Nero over 1 year

    Looking for a solution the following integral. With $n \geq2$, $\alpha>1$, $$z(n,\alpha)=\frac{\left(\alpha (n-1)\right)^n}{\Gamma (n)-\Gamma (n,(n-1) \alpha )} \int_1^\infty (y-1)^{n-1} y^{-n} e^{(\alpha -\alpha n)\frac{(y-1) }{y}} \; dy $$

    I am adding the numerical integration for different values of $\alpha$. I am adding the numerical integration for different values of $\alpha$.

    I am showing how the integral behaves (for $n=5$ and $\alpha = 3/2$ as one answer was that it does not converge: enter image description here.

    • Lucian
      Lucian about 8 years
      Let $t=\dfrac1y$ and check if the new integral converges.
    • Nero
      Nero about 8 years
      I get $\int_0^1 \frac{\exp \left(\left(\frac{1}{t}-1\right) t (\alpha -\alpha n)+(n-1) \log \left(\frac{1}{t}-1\right)-n \log \left(\frac{1}{t}\right)\right)}{t^2} \, dt$ which is still tough. Thanks.
    • Seow Fan Chong
      Seow Fan Chong about 8 years
      Let z = ln(y). Then let w = exp(-z).
    • BCLC
      BCLC about 8 years
  • Nero
    Nero about 8 years
    Integrating numerically, it converges. For instance, with $n=10^4$ and $\alpha=3/2$, I get $z(n,\alpha)=3.01$.
  • Ron Gordon
    Ron Gordon about 8 years
    @Nero: how on earth? Are you avoiding the $u=0$ endpoint? This is where the integral diverges.
  • Nero
    Nero about 8 years
    I did $z(n,\alpha)$ the original integral above, from 1 to infinity. I added a graph.
  • Ron Gordon
    Ron Gordon about 8 years
    @Nero: Then look at my integral and tell me where I went wrong. Because if I am not wrong, your numerical results are.
  • Nero
    Nero about 8 years
    I redid your transformation; it is ok (except that you missed a negative sign in front of $e^{-\alpha(n-1)}$); your integral converges numerically.
  • Ron Gordon
    Ron Gordon about 8 years
    @Nero: even if I got a sign wrong, I don't see how a nonconvergent integral "converges numerically." Either I am wrong or you are. But the way I see things, the integral diverges logarithmically near the $u=0$ endpoint. That is, the integrand has a nonintegrable singularity. No numerical method will relieve this. Now there may be a Cauchy principal value of this integral, and maybe your numerical result reflects that, maybe not. But this integral, as stated, diverges.
  • Nero
    Nero about 8 years
    I added a graph of the left side (the original integral) for values in the range [1, 100] in my original question.
  • Nero
    Nero about 8 years
    I see what happens. The integral doesn't converge in principle, but does in practive. the tail to the right doesn't fall to 0, but to a very small value.
  • Ron Gordon
    Ron Gordon about 8 years
    @Nero: I don't even know what that means. A non-convergent integral doesn't "practically" converge, or whatever adjective you want to apply. Whatever number you are getting as a result of a numerical approximation is not trustworthy. The tail of the integrand falls off as $1/u$, thus the integral diverges. If as you say the integrand does not fall off to zero then the integral diverges still more. I do not know what more I can tell you.
  • Lord_Farin
    Lord_Farin about 8 years
    This appears overly short as an answer. Please consider editing it to include an explanation of your observations, or converting it to a comment.