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Does the following integral converge or not? \begin{align} && \sum_{k=0}^{\infty} (-\varphi)^k \binom{\frac1\varphi+k}{k}\int_{-\infty}^\infty\beta x^n e^{-\beta x(k+1)}dx&& \end{align} where $e,\beta$ and $\varphi>0$ and $n$ is a positive integer.

Well I just simplified the following integral png and I got the above integral.

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  • $\begingroup$ What did you try? Does the domain of integration really start at $-\infty$? $\endgroup$
    – Tobias
    Mar 28, 2014 at 3:48
  • $\begingroup$ yes the domain starts at - infinity. $\endgroup$
    – SA-255525
    Mar 28, 2014 at 3:53
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    $\begingroup$ Then it is quite simple. Why did I ask that? (Note, the integrals must converge first else the sum is undefined.) $\endgroup$
    – Tobias
    Mar 28, 2014 at 4:18
  • $\begingroup$ what transformation/change do you suggest to make the integral converge? $\endgroup$
    – SA-255525
    Mar 28, 2014 at 10:04
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    $\begingroup$ @SA-255525 If the lower limit of integration is $-\infty$, then the answer to your question is very simple: since $\int_{-\infty}^\infty\beta x^n e^{-\beta x(k+1)}dx$ diverges, $\sum_{k=0}^{\infty} (-\varphi)^k \binom{\frac1\varphi+k}{k}\int_{-\infty}^\infty\beta x^n e^{-\beta x(k+1)}dx=\sum_{k=0}^{\infty} (-\varphi)^k \binom{\frac1\varphi+k}{k}\infty$ automatically diverges as well. A necessary condition for convergence is that the lower limit of integration be finite. $\endgroup$
    – David H
    Jun 13, 2014 at 3:31

2 Answers 2

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$$ \beta \frac{x^n e^{-\beta x}} {\left(1+\varphi e^{-\beta x}\right)^{\frac{1}{\varphi}+1}}\approx \frac{\beta}{\varphi^{1+\varphi^{-1}}} x^n \exp(\frac{\beta}{\varphi}x) \quad\mbox{ for } x\to-\infty, $$ so it is clearly convergent at this end, and $$ \beta \frac{x^n e^{-\beta x}} {\left(1+\varphi e^{-\beta x}\right)^{\frac{1}{\varphi}+1}}\approx \beta x^n \exp(- \beta x) \quad\mbox{ for } x\to \infty, $$ so it is also clearly convergent at the other end, so it is indeed convergent.

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  • $\begingroup$ @ Athanagor Wurlitzer , I am not sure how you got exp((b/phi)x). $\endgroup$
    – SA-255525
    Jun 18, 2014 at 17:37
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    $\begingroup$ @SA-255525 when $x\ll-1$, $1$ is negligible compared to $\exp(-\beta x)\phi$ which is very large. So the denominator is $(\phi \exp(-\beta x))^{1+\phi^{-1}}$, which is $\phi^{1+\phi^{-1}}\exp(-\beta/\phi x -\beta x)$. Then simplify with numerator $\endgroup$
    – user145413
    Jun 18, 2014 at 23:09
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Consider $\int_{-\infty}^\infty e^{-\beta x}\left(1+\varphi e^{-bx}\right)^{-\frac{1}{\varphi}-1}~dx$ ,

$\int_{-\infty}^\infty e^{-\beta x}\left(1+\varphi e^{-bx}\right)^{-\frac{1}{\varphi}-1}~dx$

$=\int_\infty^0x^\frac{\beta}{b}~(1+\varphi x)^{-\frac{1}{\varphi}-1}~d\left(-\dfrac{\ln x}{b}\right)$

$=\dfrac{1}{b}\int_0^\infty x^{\frac{\beta}{b}-1}(1+\varphi x)^{-\frac{1}{\varphi}-1}~dx$

$=\dfrac{1}{b}\int_0^\infty\left(\dfrac{x}{\varphi}\right)^{\frac{\beta}{b}-1}(1+x)^{-\frac{1}{\varphi}-1}~d\left(\dfrac{x}{\varphi}\right)$

$=\dfrac{1}{\varphi^\frac{\beta}{b}~b}\int_0^\infty x^{\frac{\beta}{b}-1}(1+x)^{-\frac{1}{\varphi}-1}~dx$

$=\dfrac{1}{\varphi^\frac{\beta}{b}~b}B\left(\dfrac{\beta}{b},\dfrac{1}{\varphi}-\dfrac{\beta}{b}+1\right)$

$\therefore\int_{-\infty}^\infty\beta x^ne^{-\beta x}\left(1+\varphi e^{-bx}\right)^{-\frac{1}{\varphi}-1}~dx=(-1)^n\beta\dfrac{d^n}{d\beta^n}\left(\dfrac{1}{\varphi^\frac{\beta}{b}~b}B\left(\dfrac{\beta}{b},\dfrac{1}{\varphi}-\dfrac{\beta}{b}+1\right)\right)(b=\beta)$

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