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While mucking about with powers of $\sin x$ and $\cos x$, I found something cool which might help me solve the following integral $$F(b,s)=\int_0^\infty \sin^b(t)e^{-st}dt$$ Integration by parts with $dv=e^{-st}dt$ gives $$F(b,s)=\frac{b}s\int_0^\infty\sin^{b-1}t\cos t\ e^{-st}dt$$ Integration by parts again with $dv=e^{-st}dt$ gives $$F(b,s)=\frac{b(1-b)}{s^2}\int_0^\infty\sin^{b-2}t\cos^2t\ e^{-st}dt+\frac{b}{s^2}\int_0^\infty\sin^bte^{-st}dt$$ $$F(b,s)=\frac{b(1-b)}{s^2}\int_0^\infty\sin^{b-2}t(1-\sin^2t)e^{-st}dt+\frac{b}{s^2}F(b,s)$$ $$F(b,s)=\frac{b(1-b)}{s^2}F(b-2,s)+\frac{b^2}{s^2}F(b,s)$$ $$\bigg(1-\frac{b^2}{s^2}\bigg)F(b,s)=\frac{b(1-b)}{s^2}F(b-2,s)$$ $$F(b,s)=\frac{b(1-b)}{s^2-b^2}F(b-2,s)$$ My question: how can we evaluate $F(b,s)$?

Update:

I've found that, for non-integer $b$, $$\int_0^\infty e^{-st}\sin^bt\ dt=\prod_{n\geq0}\frac{(b-2n)(1-b+2n)}{s^2-(b-2n)^2}$$ $$\int_0^\infty e^{-st}\cos^bt\ dt=\prod_{n\geq0}\frac{(b-2n)(b-2n-1)}{s^2+(b-2n)^2}$$ I'm still working on the case in which $b$ is an integer, though.

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    $\begingroup$ Well, $F(1,s)=\frac{1}{s^2+1}$: lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html , so you now can find $F(b,s)$ for all odd positive integer $b$. $\endgroup$
    – Andy Walls
    Nov 8 '18 at 3:31
  • $\begingroup$ @AndyWalls how do we use $F(1,s)$ to find $F(b,s)$ for odd natural $b$? $\endgroup$
    – clathratus
    Nov 8 '18 at 3:34
  • $\begingroup$ You can find $F(3, s)$ using $F(1, s)$. You can find $F(5, s)$ using $F(3, s)$, etc. $\endgroup$
    – Andy Walls
    Nov 8 '18 at 3:36
  • $\begingroup$ @andywalls right... can we use it to find the even values? $\endgroup$
    – clathratus
    Nov 8 '18 at 3:38
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    $\begingroup$ Find the convolution of $F(1, s)$ with itself to find $F(2, s)$, I think. Or use the integral directly to find $F(2, s)$ using a trig identity to reduce $\sin^2()$ to something simpler. $\endgroup$
    – Andy Walls
    Nov 8 '18 at 3:47
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$F(b,s)$ can be calculated for integer values of $b$ using recursion: $$F(b,s)=\frac{b}{s}\int_0^\infty \sin^{b-1}t \cos{t}e^{-st}\,dt=-\frac{b}{s^2}\int_0^\infty \sin^{b-1}t \cos{t}\,d(e^{-st})$$ $$=\frac{b}{s^2}\int_0^\infty ((b-1)\sin^{b-2}t \cos^2{t}-\sin^b{t})e^{-st}\, dt$$ $$=\frac{b}{s^2}\int_0^\infty ((b-1)\sin^{b-2}t-b\sin^b{t})e^{-st}\, dt=\frac{b(b-1)}{s^2}F(b-2,s)-\frac{b^2}{s^2}F(b,s),$$ which leads to: $$F(b,s)=\frac{b(b-1)}{b^2+s^2}F(b-2,s)$$ Also, $F(0,s)=1/s$ (integral of the exponential), and $F(1,s)=1/(1+s^2)$ (can be obtained by representing $\sin{x}$ as a sum of exponentials). Therefore: $$F(b,s)=b!s\,\Pi_{k=0}^{\lfloor b/2\rfloor}\frac{1}{s^2+(2k)^2}, \quad b=2n$$ $$F(b,s)=b!\,\Pi_{k=0}^{\lfloor b/2\rfloor}\frac{1}{s^2+(2k+1)^2}, \quad b=2n-1$$

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  • $\begingroup$ Beautiful! One question: does the $d(e^{-st})$ mean integration by parts with $dv=e^{-st}dt$? I am unfamiliar with this notation. $\endgroup$
    – clathratus
    Nov 9 '18 at 16:09
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    $\begingroup$ $d$ is the differential of a function: if $v=e^{-st}$, then $dv=d(e^{-st})=-se^{-st}\,dt$ $\endgroup$
    – atarasenko
    Nov 9 '18 at 20:15
  • $\begingroup$ Oh now I see. Thanks! $\endgroup$
    – clathratus
    Nov 9 '18 at 23:14

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