So for the function $f(x) = \exp(-x)$ I have the formula for the coefficients of

$$f(x) = \sum_{n=0}^{\infty}a_n P_n(x)$$

which is(by using Rodrigues formula)

$$a_n = \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^ndx$$

I think my next step is the integrate by parts,

$$a_n = \frac{2n+1}{2} -[\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n]^{1}_{-1} + \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx = \frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx $$

Since the term in the brackets is zero.

It seems to me that if we do this $n$ times (further $n-1$ times). We obtain.

$$a_n = \frac{2n+1}{2}\int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}(x^2-1)^ndx$$

I think I have an idea on how to proceed.


Let $$\alpha_n = \int_{-1}^{1}\exp(-x)(x^2-1)^ndx$$

Then $$\alpha_n = 2n\int_{-1}^{1}\exp(-x)x(x^2-1)^{n-1}dx = $$ $$-2n(n-1)\alpha_{n-1} - 2n(n-1)\alpha_{n-2} - 2n(n-1)\int_{-1}^{1}\exp(-x)x(x^2-1)^{n-2}dx$$

I can not seem to rid the $x$ in the last integral.

I have integrated $xe^{-x}$ before integrating the second time then added and subtracted $\alpha_{n-2}$

  • $\begingroup$ You can evaluate $\int_{-1}^{1}e^{-x}x^{k}dx$ for any $k$ using integration by parts. It may not be so elegant or clean, but you do get an answer. $\endgroup$ May 9, 2016 at 0:14
  • $\begingroup$ @HMPARTICLE Is it your question to prove the following? $$\frac{2n+1}{2}\int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}(x^2-1)^ndx=\frac{2n+1}{2} \int_{-1}^{1}\exp(-x)\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^ndx$$ $\endgroup$ May 26, 2016 at 20:29
  • $\begingroup$ Sorry, no, my question is not worded great, it is to get a recurrence relation in $\alpha_n$ $\endgroup$
    – user197848
    May 27, 2016 at 18:00

2 Answers 2


To derive a recurrence relation for such coefficients is the same as finding the structure of the continued fraction of $e^2$ (explanation at the bottom of this answer): that can be done by exploiting the Gauss continued fraction for the hyperbolic cotangent, but it is easier to compute such coefficients directly.

If we start from $$ g(t)=\int_{-1}^{1}\frac{e^{-x}}{\sqrt{1+t^2-2t x}}\,dx = \sum_{n\geq 0}\left(\int_{-1}^{1}P_n(x)e^{-x}\,dx \right)t^n\tag{1} $$ we may recover our coefficients through the Taylor series of $g(t)$ at $t=0^+$.

If we define $F(x)$ as the Dawson integral $$ F(x)=e^{-x^2}\int_{0}^{x}e^{y^2}\,dy \tag{2}$$ we have: $$ g(t) = \sqrt{\frac{2}{t}}\left[e\cdot F\left(\frac{1+t}{\sqrt{2t}}\right)-e^{-1}\cdot F\left(\frac{1-t}{\sqrt{2t}}\right)\right] \tag{3} $$ but the asymptotic expansion of $F(x)$ for large $x$ is given by: $$ F(x) = \sum_{n\geq 0}\frac{(2n-1)!!}{2^{n+1} x^{2n+1}}\tag{4} $$ hence: $$\begin{eqnarray*}[t^n]\,g(t) &=& [t^n]\left[e\cdot\sum_{m\geq 0}\frac{(2n-1)!! t^m}{(1+t)^{2m+1}}-e^{-1}\cdot\sum_{m\geq 0}\frac{(2n-1)!! t^m}{(1-t)^{2m+1}}\right]\end{eqnarray*}\tag{5}$$ but due to stars and bars we have: $$ [t^n]\frac{t^m}{(1-t)^{2m+1}} = \binom{m+n}{2m},\qquad [t^n]\frac{t^m}{(1+t)^{2m+1}}=\binom{m+n}{2m}(-1)^{n+m} \tag{6}$$ as soon as $m\leq n$, hence by $(5)$ and $(6)$ we have:

$$ \int_{-1}^{1}P_n(x)\,e^{-x}\,dx =\color{red}{\sum_{m=0}^{n}(2n-1)!!\binom{n+m}{n-m}\left(e-\frac{(-1)^{m+n}}{e}\right)}\tag{7}$$

where the RHS is a linear combination of $e$ and $e^{-1}$ with integer coefficients. This has a very interesting consequence: since $e^{-x}$ is a $C^{\infty}$ function over $(-1,1)$, the coefficients $$ c_n = \int_{-1}^{1}P_n(x)\,e^{-x}\,dx $$ decay very fast to zero as $n\to +\infty$ by the Riemann-Lebesgue/Paley-Wiener theorem, so $(7)$ provides extremely good rational approximation of $e^2$. For instance: $$ \int_{-1}^{1}P_4(x)e^{-x}\,dx = 36e-\frac{266}{e}\quad\Longrightarrow\quad e^2\approx\color{red}{\frac{133}{18}}$$ where the approximation error is $\leq 1.673\cdot 10^{-4}$. With $n=5$ we get $e^2\approx\color{red}{\frac{2431}{329}}$ whose approximation error is $\leq 1.652\cdot 10^{-6}$. And these approximation are so good that they are convergents of the continued fraction of $e^2$ by Lagrange's theorem.

Now back to the original question about the recursion. Since $$(x^2-1) P_n'(x) = nx P_n(x)-nP_{n-1}(x),\qquad P_{n+1}'(x)-P_{n-1}'(x)=(2n+1)P_n(x)\tag{8}$$ and $$ c_n = \int_{-1}^{1}e^{-x}P_n(x) = \left. -e^{-x}P_n(x)\right|_{-1}^{1}+\int_{-1}^{1}e^{-x} P_n'(x)\,dx \tag{9} $$ we have:

$$ \color{red}{c_{n+1}-c_{n-1} = (2n+1)\,c_n} \tag{10} $$ since $P_n(1)=1$ and $P_n(-1)=(-1)^n$.

Now you may prove $(7)$ through $(10)+\text{induction}$, if you like it.

In the meanwhile, I hope you discovered many interesting facts.


You have noted that \begin{align} a_n & = \frac{2n+1}{2}\int_{-1}^{1}e^{-x}\frac{1}{2^n n!}\frac{d^{n}}{dx^{n}}(x^2-1)^{n}dx \\ & = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}\frac{d^{n}}{dx^{n}}(x^2-1)^ndx \end{align} And, \begin{align} \int fg^{(n)}dx&=fg^{(n-1)}-f^{(1)}f^{(n-2)}+f^{(2)}{g^{(n-3)}}-\cdots+(-1)^{n}\int f^{(n)}gdx \end{align} Because $(x^2-1)^n=(x-1)^n(x+1)^n$, then all derivatives of $(x^2-1)$ of order $k < n$ vanish at the endpoints of $[-1,1]$. So, using a standard binomial expansion of $(x^2-1)^n$, \begin{align} a_n & = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}(x^2-1)^ndx \\ & = \frac{2n+1}{2^{n+1}n!}\int_{-1}^{1}e^{-x}\sum_{k=0}^{n}{{n}\choose{k}}(-1)^{k}x^{2k}dx \\ & = \frac{2n+1}{2^{n+1}n!}\sum_{k=0}^{n}{{n}\choose{k}}\int_{-1}^{1}e^{-x}x^{2k}dx \end{align} Then \begin{align} \int e^{-x}x^{2k}dx & = -e^{-x}x^{2k}+\int e^{-x}2kx^{2k-1}dx \\ & = -e^{-x}x^{2k}-e^{-x}(2k)x^{2k-1}+\int e^{-x}(2k)(2k-1)x^{2k-2}dx \\ & = -e^{-x}\{ x^{2k}+(2k)x^{2k-1}+(2k)(2k-1)x^{2k-2}+\cdots+(2k)! \} \end{align} It's ugly, but you do get the coefficients once you evaluate the above from $-1$ to $1$.


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