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Is there a formula to calculate the Laplace transform of $\sin^n(at)$? Being n a positive integer and a a real number.

Thank you.

AldousVexPort
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2 Answers2

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Gradshteyn and Ryzhik, in their "Table of Integrals, Series and Products", Academic Press, 2nd printing 1981, give page 478 the following two formulas according to the parity of the exponent:

Formula 3.895.1:

$$\int_0^{\infty}e^{-sx}\sin^{2m}(x)dx=\dfrac{(2m)!}{s(s^2+2^2)(s^2+4^2) \cdots (s^2+(2m)^2)}$$

Formula 3.895.5:

$$\int_0^{\infty}e^{-sx}\sin^{2m+1}(x)dx=\dfrac{(2m+1)!(1+e^{-s\pi})}{(s^2+1^2)(s^2+3^2) \cdots (s^2+(2m+1)^2)}$$

A simple change of variable will give the Laplace Transforms for $\sin^n(ax).$

Analogous formulas for $cos^n(x)$:

Formula 3.895.7:

$$\int_0^{\infty}e^{-sx}\cos^{2m}(x)dx=\dfrac{(2m)!}{s(s^2+2^2)(s^2+4^2) \cdots (s^2+(2m-2)^2)} \times$$ $$\times \left(1+\dfrac{s^2}{2!}+\dfrac{s^2(s^2+2^2)}{4!}+\cdots+\dfrac{s^2(s^2+2^2)\cdots (s^2+(2m)^2) }{(2m)!}\right)$$

Formula 3.895.9:

$$\int_0^{\infty}e^{-sx}\cos^{2m+1}(x)dx=\dfrac{(2m+1)!s}{(s^2+1^2)(s^2+3^2) \cdots (s^2+(2m+1)^2)} \times$$ $$\times \left(1+\dfrac{s^2+1}{3!}+\dfrac{(s^2+1^2)(s^2+3^2)}{5!}+\cdots+\dfrac{(s^2+1^2)(s^2+3^2)\cdots (s^2+(2m-1)^2) }{(2m+1)!}\right)$$

Jean Marie
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It is possible to write $\sin^{n}{(ax)}$ as a polynomial in $\sin{(ax)}$, so you could find out about the corresponding coefficients and perform the sum.

You can also integrate by parts to get a recurrence formula. Do check the following but I think it is

$$ Lf_{n}(u)=Lf_{n-2}(u)-\left(\frac{1}{n-1}+\frac{u^{2}}{a^{2}n(n-1)}\right)Lf_{n}(u) $$ where $Lf(u)=\int_{0}^{\infty}e^{-ux}f(x)dx$ and $f_{n}(x)=\sin^{n}{(ax)}$

shdp
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