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I'm successively integrating $x^{n} \cos{k x}$ for increasing values of positive integer n. I'm finding:



$\frac{2 x \cos{kx}}{k^2}+\frac{\left(-2+k^2 x^2\right)sin{kx}}{k^3}$,

$\frac{3 \left(-2+k^2 x^2\right) \cos{kx}}{k^4}+\frac{x \left(-6+k^2 x^2\right) \sin(kx)}{k^3}$

Is there a name for the sequence of polynomials: $x$, $2x$, $k^2x^2-2$, $3(k^2x^2-2)$, $x(k^2x^2-6)$ ... ?

Here is more:


$\frac{\cos{kx}}{k^2}+\frac{x \sin{kx}}{k}$

$\frac{2 x \cos{kx}}{k^2}+\frac{\left(-2+k^2 x^2\right) \sin{kx}}{k^3}$

$\frac{3 \left(-2+k^2 x^2\right) \cos{kx}}{k^4}+\frac{x \left(-6+k^2 x^2\right) \sin{kx}}{k^3}$

$\frac{4 x \left(-6+k^2 x^2\right) \cos{kx}}{k^4}+\frac{\left(24-12 k^2 x^2+k^4 x^4\right) \sin{kx}}{k^5}$

$\frac{5 \left(24-12 k^2 x^2+k^4 x^4\right) \cos{kx}}{k^6}+\frac{x \left(120-20 k^2 x^2+k^4 x^4\right) \sin{kx}}{k^5}$

$\frac{6 x \left(120-20 k^2 x^2+k^4 x^4\right) \cos{kx}}{k^6}+\frac{\left(-720+360 k^2 x^2-30 k^4 x^4+k^6 x^6\right) \sin{kx}}{k^7}$

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They look a bit like Hermite polynomials. –  Antonio Vargas Jun 4 '12 at 16:30
They don't seem to be exactly Hermite polynomials :-) –  Frank Jun 4 '12 at 17:41
Indeed, but for $k=2$ it might be a good place to start your search. –  Antonio Vargas Jun 4 '12 at 18:29
Is there a website for sequences of polynomials like OEIS: oeis.org? –  Frank Jun 4 '12 at 19:07
I did some search but didn't turn up anything. It doesn't seem to be an interesting sequence of polynomials, even for k = 2. –  Frank Jun 4 '12 at 19:17

2 Answers 2

The polynomials are recursive in nature, and this behavior is most apparent when the given integral

$$ \int x^n \cos(k x)\,dx $$

is viewed as the real part of the function

$$ F_{n,k}(x) = \int x^n e^{i k x}\,dx = \int x^n \cos(k x)\,dx + i\!\int x^n \sin(k x)\,dx. $$

Using integration by parts twice we can derive an inhomogeneous recurrence relation for these functions,

$$ k^2 F_{n,k}(x) - i k (n-1) F_{n-1,k}(x) + (n-1) F_{n-2,k}(x) = e^{i k x}\left(x^{n-1}-i k x^n\right). \tag{1} $$

Here we can define the polynomials

$$ P_{n,k}(x) = k^{n+1} e^{-i k x} F_{n,k}(x), $$

and then use $(1)$ to derive their recurrence relation

$$ P_{n,k}(x) - i (n-1) P_{n-1,k}(x) + (n-1) P_{n-2,k}(x) = (kx)^{n-1} - i (kx)^n. \tag{2} $$

Now, if we write

$$ k^{n+1} \int x^n \cos(k x)\,dx = A_{n,k}(x) \cos(kx) - B_{n,k}(x) \sin(kx), $$

where $A_{n,k}(x)$ and $B_{n,k}(x)$ are polynomials, we have

$$ A_{n,k}(x) = \operatorname{Re} P_{n,k}(x) \qquad \text{and} \qquad B_{n,k}(x) = \operatorname{Im}\, P_{n,k}(x). $$

The first few polynomials $P_{n,k}(x)$ are

$$ \begin{align} P_{1,k}(x) &= 1-i k x \\ P_{2,k}(x) &= 2 i+2 k x-i k^2 x^2 \\ P_{3,k}(x) &= -6+6 i k x+3 k^2 x^2-i k^3 x^3 \\ P_{4,k}(x) &= -24 i-24 k x+12 i k^2 x^2+4 k^3 x^3-i k^4 x^4 \\ P_{5,k}(x) &= 120-120 i k x-60 k^2 x^2+20 i k^3 x^3+5 k^4 x^4-i k^5 x^5 \\ P_{6,k}(x) &= 720 i+720 k x-360 i k^2 x^2-120 k^3 x^3+30 i k^4 x^4+6 k^5 x^5-i k^6 x^6 \\ P_{7,k}(x) &= -5040+5040 i k x+2520 k^2 x^2-840 i k^3 x^3-210 k^4 x^4+42 i k^5 x^5+7 k^6 x^6-i k^7 x^7 \end{align} $$

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Thanks for spending the time to look at this. –  Frank Jun 5 '12 at 2:28
Did you mean that $F_{n,k}(x) = x^{n} e^{ikx}$? –  Frank Jun 5 '12 at 2:52
@Frank, Sure thing :). And no, I meant the integral of that. Or did I make a typo somewhere? –  Antonio Vargas Jun 5 '12 at 3:09
Not sure... $F_{n,k}(x)=\int_{0}^{x} u^n e^{i k u}\,du$ ? –  Frank Jun 5 '12 at 3:21
@Frank, I meant $\int x^n e^{i k x}\,dx$ as an indefinite integral, evaluated by repeated integration by parts (keep differentiating the $x^n$ and integrating the $e^{ikx}$), so that $$\begin{align} \int e^{ikx}\,dx &= -\frac{ie^{ikx}}{k}, \\ \int x e^{ikx}\,dx &= \frac{e^{i k x} (1-i k x)}{k^2}, \\ \int x^2 e^{ikx}\,dx &= \frac{e^{i k x} \left(2 i+2 k x-i k^2 x^2\right)}{k^3}, \\ \int x^3 e^{ikx}\,dx &= \frac{e^{i k x} \left(-6+6 i k x+3 k^2 x^2-i k^3 x^3\right)}{k^4},\end{align}$$ etc. –  Antonio Vargas Jun 5 '12 at 3:47

You could also consider the exponential generating function $$ \sum_{n=0}^\infty \dfrac{t^n}{n!} \int_0^x s^n e^{iks}\ ds = \int_0^x e^{(t+ik)s}\ ds = \dfrac{e^{(t+ik)x} - 1}{t+ik}$$ This is the product of $e^{(t+ik)x}-1 = -1 + \sum_{n=0}^\infty \dfrac{t^n x^n}{n!} e^{ikx}$ and $\dfrac{1}{t+ik} = \sum_{n=0}^\infty (-1)^n \dfrac{t^n}{(ik)^{n+1}}$, so $$\int_0^x s^n e^{iks}\ ds = n! \left(\dfrac{1}{(-ik)^{n+1}} + \sum_{j=0}^n e^{ikx} \dfrac{(-1)^{n-j}}{j! (ik)^{n-j+1}} x^j\right)$$

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