integrating real powers of the cosine function using the floor function and reduction formula












1












$begingroup$


I am wondering if there is any closed form expression for
$$int cos(theta)^{m} mathrm dtheta,$$ where $minmathbb{R}$. I believe the reduction formula



$$int cos(theta)^{m} mathrm dtheta=frac{1}{m}sin(theta)cos(theta)^{m-1} + c + frac{m-1}{m}int cos(theta)^{m-2} mathrm dtheta$$



only holds for $minmathbb{Q}$. Thus I am wondering if there is a solution by splitting the power into an integer part and a decimal part using the floor function $lfloor m rfloor$ as follows:
$$
cos(theta)^{m}=cos(theta)^{(lfloor m rfloor+d)},
$$
where $d=m-lfloor m rfloor$, which gives
$$
int cos(theta)^{m}mathrm dtheta=int cos(theta)^{lfloor m rfloor}cos(theta)^{d} mathrm dtheta.
$$



Presumably I can use integration by parts on this as long as I choose to integrate $cos(theta)^{lfloor m rfloor}$ since I can use the reduction formula on this?






integration trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you do not restrict the integration bounds, $cos theta$ will have negative values and non-integer powers will be difficult to handle (even if you allow complex values for the integral).
    $endgroup$
    – gammatester
    Jan 14 '15 at 9:06












  • $begingroup$
    Thanks @gammatester for your reply. I am only interested in lower and upper limits of integration of the form $L:=((2k+1)pi)/2$ and $U:=((2k+1)pi)/2 + pi$ for $kinmathbb{Z}$, which ensures $cos(theta)$ is positive in these ranges. Forgive my ignorance but why will non-integer powers lead to complex values for the integral?
    $endgroup$
    – dandar
    Jan 14 '15 at 13:07










  • $begingroup$
    For example: Using principal values you have $(cos pi)^frac{1}{2} = (-1)^ frac{1}{2}=sqrt{-1}= i.$
    $endgroup$
    – gammatester
    Jan 14 '15 at 13:25








  • 1




    $begingroup$
    Here another example with your bounds using $k=0$ and $m=frac{3}{2}.$ Maple gives the following result with the elliptic integral $K:$ $$int_{pi/2}^{pi/2 + pi} (cos theta)^{3/2} d theta = -frac{2}{3}isqrt{2}Kleft(frac{sqrt{2}}{2}right) approx -1.748 i$$
    $endgroup$
    – gammatester
    Jan 14 '15 at 13:42












  • $begingroup$
    Thanks again @gammatester - very useful examples. However I am still not sure why your second example is complex when $cos(theta)$ is positive in that range. Can this be shown analytically as to why this is so? Getting back to my original question I presume not because I suspect there is no closed form solution to these integrals when $m$ is not an integer.
    $endgroup$
    – dandar
    Jan 14 '15 at 17:10
















1












$begingroup$


I am wondering if there is any closed form expression for
$$int cos(theta)^{m} mathrm dtheta,$$ where $minmathbb{R}$. I believe the reduction formula



$$int cos(theta)^{m} mathrm dtheta=frac{1}{m}sin(theta)cos(theta)^{m-1} + c + frac{m-1}{m}int cos(theta)^{m-2} mathrm dtheta$$



only holds for $minmathbb{Q}$. Thus I am wondering if there is a solution by splitting the power into an integer part and a decimal part using the floor function $lfloor m rfloor$ as follows:
$$
cos(theta)^{m}=cos(theta)^{(lfloor m rfloor+d)},
$$
where $d=m-lfloor m rfloor$, which gives
$$
int cos(theta)^{m}mathrm dtheta=int cos(theta)^{lfloor m rfloor}cos(theta)^{d} mathrm dtheta.
$$



Presumably I can use integration by parts on this as long as I choose to integrate $cos(theta)^{lfloor m rfloor}$ since I can use the reduction formula on this?






integration trigonometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you do not restrict the integration bounds, $cos theta$ will have negative values and non-integer powers will be difficult to handle (even if you allow complex values for the integral).
    $endgroup$
    – gammatester
    Jan 14 '15 at 9:06












  • $begingroup$
    Thanks @gammatester for your reply. I am only interested in lower and upper limits of integration of the form $L:=((2k+1)pi)/2$ and $U:=((2k+1)pi)/2 + pi$ for $kinmathbb{Z}$, which ensures $cos(theta)$ is positive in these ranges. Forgive my ignorance but why will non-integer powers lead to complex values for the integral?
    $endgroup$
    – dandar
    Jan 14 '15 at 13:07










  • $begingroup$
    For example: Using principal values you have $(cos pi)^frac{1}{2} = (-1)^ frac{1}{2}=sqrt{-1}= i.$
    $endgroup$
    – gammatester
    Jan 14 '15 at 13:25








  • 1




    $begingroup$
    Here another example with your bounds using $k=0$ and $m=frac{3}{2}.$ Maple gives the following result with the elliptic integral $K:$ $$int_{pi/2}^{pi/2 + pi} (cos theta)^{3/2} d theta = -frac{2}{3}isqrt{2}Kleft(frac{sqrt{2}}{2}right) approx -1.748 i$$
    $endgroup$
    – gammatester
    Jan 14 '15 at 13:42












  • $begingroup$
    Thanks again @gammatester - very useful examples. However I am still not sure why your second example is complex when $cos(theta)$ is positive in that range. Can this be shown analytically as to why this is so? Getting back to my original question I presume not because I suspect there is no closed form solution to these integrals when $m$ is not an integer.
    $endgroup$
    – dandar
    Jan 14 '15 at 17:10














1












1








1





$begingroup$


I am wondering if there is any closed form expression for
$$int cos(theta)^{m} mathrm dtheta,$$ where $minmathbb{R}$. I believe the reduction formula



$$int cos(theta)^{m} mathrm dtheta=frac{1}{m}sin(theta)cos(theta)^{m-1} + c + frac{m-1}{m}int cos(theta)^{m-2} mathrm dtheta$$



only holds for $minmathbb{Q}$. Thus I am wondering if there is a solution by splitting the power into an integer part and a decimal part using the floor function $lfloor m rfloor$ as follows:
$$
cos(theta)^{m}=cos(theta)^{(lfloor m rfloor+d)},
$$
where $d=m-lfloor m rfloor$, which gives
$$
int cos(theta)^{m}mathrm dtheta=int cos(theta)^{lfloor m rfloor}cos(theta)^{d} mathrm dtheta.
$$



Presumably I can use integration by parts on this as long as I choose to integrate $cos(theta)^{lfloor m rfloor}$ since I can use the reduction formula on this?






integration trigonometry







share|cite|improve this question











$endgroup$




I am wondering if there is any closed form expression for
$$int cos(theta)^{m} mathrm dtheta,$$ where $minmathbb{R}$. I believe the reduction formula



$$int cos(theta)^{m} mathrm dtheta=frac{1}{m}sin(theta)cos(theta)^{m-1} + c + frac{m-1}{m}int cos(theta)^{m-2} mathrm dtheta$$



only holds for $minmathbb{Q}$. Thus I am wondering if there is a solution by splitting the power into an integer part and a decimal part using the floor function $lfloor m rfloor$ as follows:
$$
cos(theta)^{m}=cos(theta)^{(lfloor m rfloor+d)},
$$
where $d=m-lfloor m rfloor$, which gives
$$
int cos(theta)^{m}mathrm dtheta=int cos(theta)^{lfloor m rfloor}cos(theta)^{d} mathrm dtheta.
$$



Presumably I can use integration by parts on this as long as I choose to integrate $cos(theta)^{lfloor m rfloor}$ since I can use the reduction formula on this?






integration trigonometry




integration trigonometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 14:43









Saad

20.3k92352




20.3k92352










asked Jan 14 '15 at 8:57









dandardandar

413516




413516












  • $begingroup$
    If you do not restrict the integration bounds, $cos theta$ will have negative values and non-integer powers will be difficult to handle (even if you allow complex values for the integral).
    $endgroup$
    – gammatester
    Jan 14 '15 at 9:06












  • $begingroup$
    Thanks @gammatester for your reply. I am only interested in lower and upper limits of integration of the form $L:=((2k+1)pi)/2$ and $U:=((2k+1)pi)/2 + pi$ for $kinmathbb{Z}$, which ensures $cos(theta)$ is positive in these ranges. Forgive my ignorance but why will non-integer powers lead to complex values for the integral?
    $endgroup$
    – dandar
    Jan 14 '15 at 13:07










  • $begingroup$
    For example: Using principal values you have $(cos pi)^frac{1}{2} = (-1)^ frac{1}{2}=sqrt{-1}= i.$
    $endgroup$
    – gammatester
    Jan 14 '15 at 13:25








  • 1




    $begingroup$
    Here another example with your bounds using $k=0$ and $m=frac{3}{2}.$ Maple gives the following result with the elliptic integral $K:$ $$int_{pi/2}^{pi/2 + pi} (cos theta)^{3/2} d theta = -frac{2}{3}isqrt{2}Kleft(frac{sqrt{2}}{2}right) approx -1.748 i$$
    $endgroup$
    – gammatester
    Jan 14 '15 at 13:42












  • $begingroup$
    Thanks again @gammatester - very useful examples. However I am still not sure why your second example is complex when $cos(theta)$ is positive in that range. Can this be shown analytically as to why this is so? Getting back to my original question I presume not because I suspect there is no closed form solution to these integrals when $m$ is not an integer.
    $endgroup$
    – dandar
    Jan 14 '15 at 17:10


















  • $begingroup$
    If you do not restrict the integration bounds, $cos theta$ will have negative values and non-integer powers will be difficult to handle (even if you allow complex values for the integral).
    $endgroup$
    – gammatester
    Jan 14 '15 at 9:06












  • $begingroup$
    Thanks @gammatester for your reply. I am only interested in lower and upper limits of integration of the form $L:=((2k+1)pi)/2$ and $U:=((2k+1)pi)/2 + pi$ for $kinmathbb{Z}$, which ensures $cos(theta)$ is positive in these ranges. Forgive my ignorance but why will non-integer powers lead to complex values for the integral?
    $endgroup$
    – dandar
    Jan 14 '15 at 13:07










  • $begingroup$
    For example: Using principal values you have $(cos pi)^frac{1}{2} = (-1)^ frac{1}{2}=sqrt{-1}= i.$
    $endgroup$
    – gammatester
    Jan 14 '15 at 13:25








  • 1




    $begingroup$
    Here another example with your bounds using $k=0$ and $m=frac{3}{2}.$ Maple gives the following result with the elliptic integral $K:$ $$int_{pi/2}^{pi/2 + pi} (cos theta)^{3/2} d theta = -frac{2}{3}isqrt{2}Kleft(frac{sqrt{2}}{2}right) approx -1.748 i$$
    $endgroup$
    – gammatester
    Jan 14 '15 at 13:42












  • $begingroup$
    Thanks again @gammatester - very useful examples. However I am still not sure why your second example is complex when $cos(theta)$ is positive in that range. Can this be shown analytically as to why this is so? Getting back to my original question I presume not because I suspect there is no closed form solution to these integrals when $m$ is not an integer.
    $endgroup$
    – dandar
    Jan 14 '15 at 17:10
















$begingroup$
If you do not restrict the integration bounds, $cos theta$ will have negative values and non-integer powers will be difficult to handle (even if you allow complex values for the integral).
$endgroup$
– gammatester
Jan 14 '15 at 9:06






$begingroup$
If you do not restrict the integration bounds, $cos theta$ will have negative values and non-integer powers will be difficult to handle (even if you allow complex values for the integral).
$endgroup$
– gammatester
Jan 14 '15 at 9:06














$begingroup$
Thanks @gammatester for your reply. I am only interested in lower and upper limits of integration of the form $L:=((2k+1)pi)/2$ and $U:=((2k+1)pi)/2 + pi$ for $kinmathbb{Z}$, which ensures $cos(theta)$ is positive in these ranges. Forgive my ignorance but why will non-integer powers lead to complex values for the integral?
$endgroup$
– dandar
Jan 14 '15 at 13:07




$begingroup$
Thanks @gammatester for your reply. I am only interested in lower and upper limits of integration of the form $L:=((2k+1)pi)/2$ and $U:=((2k+1)pi)/2 + pi$ for $kinmathbb{Z}$, which ensures $cos(theta)$ is positive in these ranges. Forgive my ignorance but why will non-integer powers lead to complex values for the integral?
$endgroup$
– dandar
Jan 14 '15 at 13:07












$begingroup$
For example: Using principal values you have $(cos pi)^frac{1}{2} = (-1)^ frac{1}{2}=sqrt{-1}= i.$
$endgroup$
– gammatester
Jan 14 '15 at 13:25






$begingroup$
For example: Using principal values you have $(cos pi)^frac{1}{2} = (-1)^ frac{1}{2}=sqrt{-1}= i.$
$endgroup$
– gammatester
Jan 14 '15 at 13:25






1




1




$begingroup$
Here another example with your bounds using $k=0$ and $m=frac{3}{2}.$ Maple gives the following result with the elliptic integral $K:$ $$int_{pi/2}^{pi/2 + pi} (cos theta)^{3/2} d theta = -frac{2}{3}isqrt{2}Kleft(frac{sqrt{2}}{2}right) approx -1.748 i$$
$endgroup$
– gammatester
Jan 14 '15 at 13:42






$begingroup$
Here another example with your bounds using $k=0$ and $m=frac{3}{2}.$ Maple gives the following result with the elliptic integral $K:$ $$int_{pi/2}^{pi/2 + pi} (cos theta)^{3/2} d theta = -frac{2}{3}isqrt{2}Kleft(frac{sqrt{2}}{2}right) approx -1.748 i$$
$endgroup$
– gammatester
Jan 14 '15 at 13:42














$begingroup$
Thanks again @gammatester - very useful examples. However I am still not sure why your second example is complex when $cos(theta)$ is positive in that range. Can this be shown analytically as to why this is so? Getting back to my original question I presume not because I suspect there is no closed form solution to these integrals when $m$ is not an integer.
$endgroup$
– dandar
Jan 14 '15 at 17:10




$begingroup$
Thanks again @gammatester - very useful examples. However I am still not sure why your second example is complex when $cos(theta)$ is positive in that range. Can this be shown analytically as to why this is so? Getting back to my original question I presume not because I suspect there is no closed form solution to these integrals when $m$ is not an integer.
$endgroup$
– dandar
Jan 14 '15 at 17:10










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