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As part of an exercise including gamma functions and Laguerre polynomials, I need to show that for a Laguerre polynomial $L_n(x)$, $$\int\limits_0^\infty L_n(x)x^ke^{-x}dx = 0 \textrm{ with } n \in \mathbb{N} \textrm{ and }k \in \{0,1,\dots,n-1\}$$ I've managed to narrow it down to proving that $$\sum\limits_{i=0}^n\frac{(-1)^{n-i}(k+i)!}{(i!)^2(n-i)!} = 0\textrm{ with } n \in \mathbb{N} \textrm{ and }k \in \{0,1,\dots,n-1\}$$ But this is where I'm stuck. I used Maple and WolframAlpha to turn this sum into $$\frac{k!(-1)^n\Gamma(n-k)}{\Gamma(-k)\Gamma(n+1)^2}$$ Which is $0$ since $\Gamma(-k)$ would be infinity, due to the Gamma function having simple poles in the negative integers. However, neither Maple or Wolfram can tell me how I convert the sum into this fraction of gamma functions. Am I missing something obvious or are there any tricks to get there?

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  • $\begingroup$ The definition as given by the exercise would be $$P_n(x) = \frac{(-1)^nn!}{2\pi i} \oint_\Sigma \frac{\Gamma(t-n)}{\Gamma(t+1)^2}x^tdt.$$ where $\Sigma$ is a closed curve that goes around the points $0, 1, \dots, n$ counterclockwise once. $\endgroup$
    – user49719
    Commented May 25, 2013 at 13:43
  • $\begingroup$ rewrite $ x^{k } $ as a linear combination of Laguerre polynomials $ L_{m} (x) $ and use the orthogonality property $ \int_{0}^{\infty} L_{m}(x)L_{n}(x)exp(-x)dx =0 $ $\endgroup$ Commented May 25, 2013 at 13:47
  • $\begingroup$ Perhaps I should've mentioned this, but Laguerre polynomials aren't actually part of my course and this exercise is where they were first introduced, so I would not be allowed to use this property without proving it first. Would it be wise to prove this property or to continue looking for a more direct solution? $\endgroup$
    – user49719
    Commented May 25, 2013 at 13:51

1 Answer 1

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With the definition you have given, \begin{align}\int_0^{\infty}L_n(x)x^ke^{-x}dx&=\frac{(-1)^n n!}{2\pi i}\oint_{\Sigma}\frac{\Gamma(t-n)}{\Gamma^2(t+1)}\left(\int_0^{\infty}x^{t+k}e^{-x}dx\right)dt=\\ &=\frac{(-1)^n n!}{2\pi i}\oint_{\Sigma}\frac{\Gamma(t-n)\Gamma(t+k+1)}{\Gamma^2(t+1)}dt \end{align} What are the singular points of the integrand?

  • $\Gamma(t-n)$ has simple poles at $t=n,n-1,n-2,\ldots$. However, almost all of these poles, namely $t=-1,-2,\ldots$ are compensated by the poles of one of the $\Gamma(t+1)$ in the denominator.

  • Also, $\Gamma(t+k+1)$ has simple poles at $-k-1,-k-2,\ldots$ but all of them are compensated by the poles of the second $\Gamma(t+1)$.

Thus we have simple poles at $t=n,n-1,\ldots,0$ - precisely those inside of $\Sigma$. But now, instead of shrinking the integration contour, let us expand it to infinity. On the circle $|t|=R$, we have (think why) $$\frac{\Gamma(t-n)\Gamma(t+k+1)}{\Gamma^2(t+1)}=O\left(R^{k-n-1}\right)\qquad \text{as}\;R\rightarrow\infty.$$ Therefore, the integral over this circle will be (at most) $O\left(R^{k-n}\right)$, but this goes to zero as $R\rightarrow\infty$ for $k<n$.

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  • $\begingroup$ Thanks! I used the same reasoning with the poles of the gamma, but turned it into a summation rather than changing the order of integration. $\endgroup$
    – user49719
    Commented May 25, 2013 at 15:11
  • $\begingroup$ Is it also possible to prove the integral will be zero using the residue formula? I'm asking this question as I'm not able to see why the integrand is $O(R^{k-n-1})$. $\endgroup$
    – 88888888
    Commented Nov 15, 2020 at 16:30
  • $\begingroup$ @88888888 What is done here is basically the residue formula, with residue zero. The ratio of gamma functions is in fact a rational function of $t$. For example, $\Gamma(t+k+1)/\Gamma(t+1)=(t+1)(t+2)\ldots(t+k)$. Likewise, $\Gamma(t+1)/\Gamma(t-n)=t(t-1)\ldots(t-n)$. The estimate $O(R^{k-n-1})$ is obtained by considering the ratio of these two polynomials. $\endgroup$ Commented Nov 15, 2020 at 17:46

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