integrating real powers of the cosine function using the floor function and reduction formula
$begingroup$
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
edited Dec 11 '18 at 14:43
Saad
20.3k92352
asked Jan 14 '15 at 8:57
dandardandar
413516
$endgroup$
|
show 1 more comment
$begingroup$
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
edited Dec 11 '18 at 14:43
Saad
20.3k92352
asked Jan 14 '15 at 8:57
dandardandar
413516
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
edited Dec 11 '18 at 14:43
Saad
20.3k92352
asked Jan 14 '15 at 8:57
dandardandar
413516
$endgroup$
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
edited Dec 11 '18 at 14:43
Saad
20.3k92352
asked Jan 14 '15 at 8:57
dandardandar
413516
edited Dec 11 '18 at 14:43
Saad
20.3k92352
asked Jan 14 '15 at 8:57
dandardandar
413516
edited Dec 11 '18 at 14:43
Saad
20.3k92352
edited Dec 11 '18 at 14:43
Saad
20.3k92352
edited Dec 11 '18 at 14:43
Saad
20.3k92352
20.3k92352
asked Jan 14 '15 at 8:57
dandardandar
413516
asked Jan 14 '15 at 8:57
dandardandar
413516
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
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$begingroup$
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