Mixing
Solving $int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx$
$begingroup$
Spurred on this question I decided to investigate the following integral:
begin{equation}
I_n = int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx
end{equation}
Where $n in mathbb{N}$.
The approach I've taken is rather simple and whilst I've arrived at a closed form solution, I'm wondering whether the resultant sum can be expressed in terms of (non)-elementary functions. Would love to see other methods that can be used to solve this! (Using any methods).
First:
begin{equation}
I_n = int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx = int_0^{frac{pi}{2}}frac{1}{left[sin^{2}(x)right]^n + left[cos^{2}(x)right]^n}:dx
end{equation}
Using the Double-Angle Formulas:
begin{equation}
sin^2(x) = frac{1 - cosleft(2xright)}{2} qquad cos^2(x) = frac{1 + cosleft(2xright)}{2}
end{equation}
Thus:
begin{align}
I_n &= int_0^{frac{pi}{2}}frac{1}{left[sin^{2}(x)right]^n + left[cos^{2}(x)right]^n}:dx = int_0^{frac{pi}{2}}frac{1}{left[frac{1 - cosleft(2xright)}{2}right]^n + left[frac{1 + cosleft(2xright)}{2}right]^n}:dx \
&= 2^n int_0^{frac{pi}{2}} frac{1}{left[1 - cosleft(2xright)right]^n + left[1 + cosleft(2xright)right]^n}:dx
end{align}
Making the substitution $u = 2x$
begin{equation}
I_n = 2^n int_0^{frac{pi}{2}} frac{1}{left[1 - cosleft(2xright)right]^n + left[1 + cosleft(2xright)right]^n}:dx = 2^{n - 1} int_0^{pi} frac{1}{left[1 - cosleft(uright)right]^n + left[1 + cosleft(uright)right]^n}:du
end{equation}
Now apply the Weierstrass (/Tangent half angle) substitution $t = tanleft(frac{u}{2}right)$:
begin{align}
I_n &= 2^{n - 1} int_0^{pi} frac{1}{left[1 - cosleft(uright)right]^n + left[1 + cosleft(uright)right]^n}:du = 2^{n - 1} int_0^{infty} frac{1}{left[1 - frac{1 - t^2}{1 + t^2}right]^n + left[1 + frac{1 - t^2}{1 + t^2}right]^n} frac{2:dt}{t^2 + 1} \
&= int_0^{infty} frac{left[1 + t^2 right]^{n - 1}}{t^{2n} + 1}:dt
end{align}
By the Binomial Theorem:
begin{equation}
left[1 + t^2 right]^{2n - 1} = sum_{j = 0}^{n - 1} {n - 1 choose j}t^{2j}
end{equation}
Thus:
begin{align}
I_n &= int_0^{infty} frac{left[1 + t^2 right]^{2n - 1}}{t^{2n} + 1}:dt = sum_{j = 0}^{n - 1} {n - 1 choose j} int_0^{infty} frac{t^{2j}}{t^{2n} + 1}
end{align}
Using the solution I found here we find:
begin{align}
I_n &= sum_{j = 0}^{n - 1} {n - 1 choose j} int_0^{infty} frac{t^{2j}}{t^{2n} + 1} = sum_{j = 0}^{n - 1} {n - 1 choose j} cdot frac{1}{2n} cdot 1^{frac{2j + 1}{2n} - 1}Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right) \
&= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right)
end{align}
Using the relationship between the Beta and Gamma function:
begin{align}
I_n &= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right) = frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} frac{Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right)}{Gammaleft(1 - frac{2j + 1}{2n} + frac{2j + 1}{2n}right)} \
&= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j}Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right)
end{align}
As $frac{2j + 1}{2n} not in mathbb{Z}$ we can employ Euler's Reflection Formula to yield:
begin{align}
I_n &= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j}Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right) = frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} frac{pi}{sinleft(pi cdot frac{2j + 1}{2n}right)} \&= frac{pi}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} operatorname{cosec}left(frac{pi}{2n}left(2j + 1right) right)
end{align}
Thus,
begin{equation}
int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx = frac{pi}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} operatorname{cosec}left(frac{pi}{2n}left(2j + 1right) right)
end{equation}
Edit - Realised that $2n - 1$ should be $n - 1$
Extra Addition:
Using the same method as above an by letting
begin{equation}
S_n(x) = sin^{2n}(x) + cos^{2n}(x)
end{equation}
It becomes rather easy to solve:
begin{equation}
I_{n,m} = int_0^{frac{pi}{2}} frac{S_n(x)}{S_m(x)}:dx
end{equation}
Where $n lt m$. After applying the double angle formulas, $u$-substitution, and half tangent formula we arrive at:
begin{equation}
I_{n,m} = int_0^{infty} frac{S_n(x)}{S_m(x)}:dx = int_0^{infty} frac{t^{2n} + 1}{t^{2m} + 1} left[1 + t^2 right]^{m - n - 1}:dt
end{equation}
Applying the Binomial Expansion:
begin{align}
I_{n,m} &= sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2n} + 1}{t^{2m} + 1} cdot t^{2j}:dt \
&= sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2left(n + jright)} }{t^{2m} + 1} :dt + sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2j} }{t^{2m} + 1} :dt\
end{align}
Again applying another of my solutions (referenced above) we arrive at:
begin{align}
I_{n,m} &= sum_{j = 0}^{n - 1} {m - n - 1 choose j} frac{pi}{2m}operatorname{cosec}left(frac{pi}{2m}left(2left(n + jright) + 1right) right) + sum_{j = 0}^{n - 1} {m - n - 1 choose j}frac{pi}{2m}operatorname{cosec}left(frac{pi}{2m}left(2j + 1right) right) \
&= frac{pi}{2m}sum_{j = 0}^{n - 1} {m - n - 1 choose j}left[operatorname{cosec}left(frac{pi}{2m}left(2left(n + jright) + 1right) right) + operatorname{cosec}left(frac{pi}{2m}left(2j + 1right) right) right]
end{align}
integration definite-integrals gamma-function trigonometric-integrals beta-function
asked Jan 3 at 2:06
DavidGDavidG
2,047720
$endgroup$
add a comment |
$begingroup$
Spurred on this question I decided to investigate the following integral:
begin{equation}
I_n = int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx
end{equation}
Where $n in mathbb{N}$.
The approach I've taken is rather simple and whilst I've arrived at a closed form solution, I'm wondering whether the resultant sum can be expressed in terms of (non)-elementary functions. Would love to see other methods that can be used to solve this! (Using any methods).
First:
begin{equation}
I_n = int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx = int_0^{frac{pi}{2}}frac{1}{left[sin^{2}(x)right]^n + left[cos^{2}(x)right]^n}:dx
end{equation}
Using the Double-Angle Formulas:
begin{equation}
sin^2(x) = frac{1 - cosleft(2xright)}{2} qquad cos^2(x) = frac{1 + cosleft(2xright)}{2}
end{equation}
Thus:
begin{align}
I_n &= int_0^{frac{pi}{2}}frac{1}{left[sin^{2}(x)right]^n + left[cos^{2}(x)right]^n}:dx = int_0^{frac{pi}{2}}frac{1}{left[frac{1 - cosleft(2xright)}{2}right]^n + left[frac{1 + cosleft(2xright)}{2}right]^n}:dx \
&= 2^n int_0^{frac{pi}{2}} frac{1}{left[1 - cosleft(2xright)right]^n + left[1 + cosleft(2xright)right]^n}:dx
end{align}
Making the substitution $u = 2x$
begin{equation}
I_n = 2^n int_0^{frac{pi}{2}} frac{1}{left[1 - cosleft(2xright)right]^n + left[1 + cosleft(2xright)right]^n}:dx = 2^{n - 1} int_0^{pi} frac{1}{left[1 - cosleft(uright)right]^n + left[1 + cosleft(uright)right]^n}:du
end{equation}
Now apply the Weierstrass (/Tangent half angle) substitution $t = tanleft(frac{u}{2}right)$:
begin{align}
I_n &= 2^{n - 1} int_0^{pi} frac{1}{left[1 - cosleft(uright)right]^n + left[1 + cosleft(uright)right]^n}:du = 2^{n - 1} int_0^{infty} frac{1}{left[1 - frac{1 - t^2}{1 + t^2}right]^n + left[1 + frac{1 - t^2}{1 + t^2}right]^n} frac{2:dt}{t^2 + 1} \
&= int_0^{infty} frac{left[1 + t^2 right]^{n - 1}}{t^{2n} + 1}:dt
end{align}
By the Binomial Theorem:
begin{equation}
left[1 + t^2 right]^{2n - 1} = sum_{j = 0}^{n - 1} {n - 1 choose j}t^{2j}
end{equation}
Thus:
begin{align}
I_n &= int_0^{infty} frac{left[1 + t^2 right]^{2n - 1}}{t^{2n} + 1}:dt = sum_{j = 0}^{n - 1} {n - 1 choose j} int_0^{infty} frac{t^{2j}}{t^{2n} + 1}
end{align}
Using the solution I found here we find:
begin{align}
I_n &= sum_{j = 0}^{n - 1} {n - 1 choose j} int_0^{infty} frac{t^{2j}}{t^{2n} + 1} = sum_{j = 0}^{n - 1} {n - 1 choose j} cdot frac{1}{2n} cdot 1^{frac{2j + 1}{2n} - 1}Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right) \
&= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right)
end{align}
Using the relationship between the Beta and Gamma function:
begin{align}
I_n &= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right) = frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} frac{Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right)}{Gammaleft(1 - frac{2j + 1}{2n} + frac{2j + 1}{2n}right)} \
&= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j}Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right)
end{align}
As $frac{2j + 1}{2n} not in mathbb{Z}$ we can employ Euler's Reflection Formula to yield:
begin{align}
I_n &= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j}Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right) = frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} frac{pi}{sinleft(pi cdot frac{2j + 1}{2n}right)} \&= frac{pi}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} operatorname{cosec}left(frac{pi}{2n}left(2j + 1right) right)
end{align}
Thus,
begin{equation}
int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx = frac{pi}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} operatorname{cosec}left(frac{pi}{2n}left(2j + 1right) right)
end{equation}
Edit - Realised that $2n - 1$ should be $n - 1$
Extra Addition:
Using the same method as above an by letting
begin{equation}
S_n(x) = sin^{2n}(x) + cos^{2n}(x)
end{equation}
It becomes rather easy to solve:
begin{equation}
I_{n,m} = int_0^{frac{pi}{2}} frac{S_n(x)}{S_m(x)}:dx
end{equation}
Where $n lt m$. After applying the double angle formulas, $u$-substitution, and half tangent formula we arrive at:
begin{equation}
I_{n,m} = int_0^{infty} frac{S_n(x)}{S_m(x)}:dx = int_0^{infty} frac{t^{2n} + 1}{t^{2m} + 1} left[1 + t^2 right]^{m - n - 1}:dt
end{equation}
Applying the Binomial Expansion:
begin{align}
I_{n,m} &= sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2n} + 1}{t^{2m} + 1} cdot t^{2j}:dt \
&= sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2left(n + jright)} }{t^{2m} + 1} :dt + sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2j} }{t^{2m} + 1} :dt\
end{align}
Again applying another of my solutions (referenced above) we arrive at:
begin{align}
I_{n,m} &= sum_{j = 0}^{n - 1} {m - n - 1 choose j} frac{pi}{2m}operatorname{cosec}left(frac{pi}{2m}left(2left(n + jright) + 1right) right) + sum_{j = 0}^{n - 1} {m - n - 1 choose j}frac{pi}{2m}operatorname{cosec}left(frac{pi}{2m}left(2j + 1right) right) \
&= frac{pi}{2m}sum_{j = 0}^{n - 1} {m - n - 1 choose j}left[operatorname{cosec}left(frac{pi}{2m}left(2left(n + jright) + 1right) right) + operatorname{cosec}left(frac{pi}{2m}left(2j + 1right) right) right]
end{align}
integration definite-integrals gamma-function trigonometric-integrals beta-function
asked Jan 3 at 2:06
DavidGDavidG
2,047720
$endgroup$
$begingroup$
I think that's the "closest" you can reach to the "closed form"
$endgroup$
– Digamma
Jan 3 at 3:54
$begingroup$
@Digamma - Thanks. Let me know if you think of anything :-)
$endgroup$
– DavidG
Jan 3 at 8:59
$begingroup$
Wikipedia defines "a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations." So if $n$ is finite, you have a closed form. Bravo David, this one is really spectacular.
$endgroup$
– clathratus
Jan 3 at 23:29
$begingroup$
Thanks @clathratus!.
$endgroup$
– DavidG
Jan 4 at 1:18
add a comment |
$begingroup$
Spurred on this question I decided to investigate the following integral:
begin{equation}
I_n = int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx
end{equation}
Where $n in mathbb{N}$.
The approach I've taken is rather simple and whilst I've arrived at a closed form solution, I'm wondering whether the resultant sum can be expressed in terms of (non)-elementary functions. Would love to see other methods that can be used to solve this! (Using any methods).
First:
begin{equation}
I_n = int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx = int_0^{frac{pi}{2}}frac{1}{left[sin^{2}(x)right]^n + left[cos^{2}(x)right]^n}:dx
end{equation}
Using the Double-Angle Formulas:
begin{equation}
sin^2(x) = frac{1 - cosleft(2xright)}{2} qquad cos^2(x) = frac{1 + cosleft(2xright)}{2}
end{equation}
Thus:
begin{align}
I_n &= int_0^{frac{pi}{2}}frac{1}{left[sin^{2}(x)right]^n + left[cos^{2}(x)right]^n}:dx = int_0^{frac{pi}{2}}frac{1}{left[frac{1 - cosleft(2xright)}{2}right]^n + left[frac{1 + cosleft(2xright)}{2}right]^n}:dx \
&= 2^n int_0^{frac{pi}{2}} frac{1}{left[1 - cosleft(2xright)right]^n + left[1 + cosleft(2xright)right]^n}:dx
end{align}
Making the substitution $u = 2x$
begin{equation}
I_n = 2^n int_0^{frac{pi}{2}} frac{1}{left[1 - cosleft(2xright)right]^n + left[1 + cosleft(2xright)right]^n}:dx = 2^{n - 1} int_0^{pi} frac{1}{left[1 - cosleft(uright)right]^n + left[1 + cosleft(uright)right]^n}:du
end{equation}
Now apply the Weierstrass (/Tangent half angle) substitution $t = tanleft(frac{u}{2}right)$:
begin{align}
I_n &= 2^{n - 1} int_0^{pi} frac{1}{left[1 - cosleft(uright)right]^n + left[1 + cosleft(uright)right]^n}:du = 2^{n - 1} int_0^{infty} frac{1}{left[1 - frac{1 - t^2}{1 + t^2}right]^n + left[1 + frac{1 - t^2}{1 + t^2}right]^n} frac{2:dt}{t^2 + 1} \
&= int_0^{infty} frac{left[1 + t^2 right]^{n - 1}}{t^{2n} + 1}:dt
end{align}
By the Binomial Theorem:
begin{equation}
left[1 + t^2 right]^{2n - 1} = sum_{j = 0}^{n - 1} {n - 1 choose j}t^{2j}
end{equation}
Thus:
begin{align}
I_n &= int_0^{infty} frac{left[1 + t^2 right]^{2n - 1}}{t^{2n} + 1}:dt = sum_{j = 0}^{n - 1} {n - 1 choose j} int_0^{infty} frac{t^{2j}}{t^{2n} + 1}
end{align}
Using the solution I found here we find:
begin{align}
I_n &= sum_{j = 0}^{n - 1} {n - 1 choose j} int_0^{infty} frac{t^{2j}}{t^{2n} + 1} = sum_{j = 0}^{n - 1} {n - 1 choose j} cdot frac{1}{2n} cdot 1^{frac{2j + 1}{2n} - 1}Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right) \
&= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right)
end{align}
Using the relationship between the Beta and Gamma function:
begin{align}
I_n &= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right) = frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} frac{Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right)}{Gammaleft(1 - frac{2j + 1}{2n} + frac{2j + 1}{2n}right)} \
&= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j}Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right)
end{align}
As $frac{2j + 1}{2n} not in mathbb{Z}$ we can employ Euler's Reflection Formula to yield:
begin{align}
I_n &= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j}Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right) = frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} frac{pi}{sinleft(pi cdot frac{2j + 1}{2n}right)} \&= frac{pi}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} operatorname{cosec}left(frac{pi}{2n}left(2j + 1right) right)
end{align}
Thus,
begin{equation}
int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx = frac{pi}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} operatorname{cosec}left(frac{pi}{2n}left(2j + 1right) right)
end{equation}
Edit - Realised that $2n - 1$ should be $n - 1$
Extra Addition:
Using the same method as above an by letting
begin{equation}
S_n(x) = sin^{2n}(x) + cos^{2n}(x)
end{equation}
It becomes rather easy to solve:
begin{equation}
I_{n,m} = int_0^{frac{pi}{2}} frac{S_n(x)}{S_m(x)}:dx
end{equation}
Where $n lt m$. After applying the double angle formulas, $u$-substitution, and half tangent formula we arrive at:
begin{equation}
I_{n,m} = int_0^{infty} frac{S_n(x)}{S_m(x)}:dx = int_0^{infty} frac{t^{2n} + 1}{t^{2m} + 1} left[1 + t^2 right]^{m - n - 1}:dt
end{equation}
Applying the Binomial Expansion:
begin{align}
I_{n,m} &= sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2n} + 1}{t^{2m} + 1} cdot t^{2j}:dt \
&= sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2left(n + jright)} }{t^{2m} + 1} :dt + sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2j} }{t^{2m} + 1} :dt\
end{align}
Again applying another of my solutions (referenced above) we arrive at:
begin{align}
I_{n,m} &= sum_{j = 0}^{n - 1} {m - n - 1 choose j} frac{pi}{2m}operatorname{cosec}left(frac{pi}{2m}left(2left(n + jright) + 1right) right) + sum_{j = 0}^{n - 1} {m - n - 1 choose j}frac{pi}{2m}operatorname{cosec}left(frac{pi}{2m}left(2j + 1right) right) \
&= frac{pi}{2m}sum_{j = 0}^{n - 1} {m - n - 1 choose j}left[operatorname{cosec}left(frac{pi}{2m}left(2left(n + jright) + 1right) right) + operatorname{cosec}left(frac{pi}{2m}left(2j + 1right) right) right]
end{align}
integration definite-integrals gamma-function trigonometric-integrals beta-function
asked Jan 3 at 2:06
DavidGDavidG
2,047720
$endgroup$
Spurred on this question I decided to investigate the following integral:
begin{equation}
I_n = int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx
end{equation}
Where $n in mathbb{N}$.
The approach I've taken is rather simple and whilst I've arrived at a closed form solution, I'm wondering whether the resultant sum can be expressed in terms of (non)-elementary functions. Would love to see other methods that can be used to solve this! (Using any methods).
First:
begin{equation}
I_n = int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx = int_0^{frac{pi}{2}}frac{1}{left[sin^{2}(x)right]^n + left[cos^{2}(x)right]^n}:dx
end{equation}
Using the Double-Angle Formulas:
begin{equation}
sin^2(x) = frac{1 - cosleft(2xright)}{2} qquad cos^2(x) = frac{1 + cosleft(2xright)}{2}
end{equation}
Thus:
begin{align}
I_n &= int_0^{frac{pi}{2}}frac{1}{left[sin^{2}(x)right]^n + left[cos^{2}(x)right]^n}:dx = int_0^{frac{pi}{2}}frac{1}{left[frac{1 - cosleft(2xright)}{2}right]^n + left[frac{1 + cosleft(2xright)}{2}right]^n}:dx \
&= 2^n int_0^{frac{pi}{2}} frac{1}{left[1 - cosleft(2xright)right]^n + left[1 + cosleft(2xright)right]^n}:dx
end{align}
Making the substitution $u = 2x$
begin{equation}
I_n = 2^n int_0^{frac{pi}{2}} frac{1}{left[1 - cosleft(2xright)right]^n + left[1 + cosleft(2xright)right]^n}:dx = 2^{n - 1} int_0^{pi} frac{1}{left[1 - cosleft(uright)right]^n + left[1 + cosleft(uright)right]^n}:du
end{equation}
Now apply the Weierstrass (/Tangent half angle) substitution $t = tanleft(frac{u}{2}right)$:
begin{align}
I_n &= 2^{n - 1} int_0^{pi} frac{1}{left[1 - cosleft(uright)right]^n + left[1 + cosleft(uright)right]^n}:du = 2^{n - 1} int_0^{infty} frac{1}{left[1 - frac{1 - t^2}{1 + t^2}right]^n + left[1 + frac{1 - t^2}{1 + t^2}right]^n} frac{2:dt}{t^2 + 1} \
&= int_0^{infty} frac{left[1 + t^2 right]^{n - 1}}{t^{2n} + 1}:dt
end{align}
By the Binomial Theorem:
begin{equation}
left[1 + t^2 right]^{2n - 1} = sum_{j = 0}^{n - 1} {n - 1 choose j}t^{2j}
end{equation}
Thus:
begin{align}
I_n &= int_0^{infty} frac{left[1 + t^2 right]^{2n - 1}}{t^{2n} + 1}:dt = sum_{j = 0}^{n - 1} {n - 1 choose j} int_0^{infty} frac{t^{2j}}{t^{2n} + 1}
end{align}
Using the solution I found here we find:
begin{align}
I_n &= sum_{j = 0}^{n - 1} {n - 1 choose j} int_0^{infty} frac{t^{2j}}{t^{2n} + 1} = sum_{j = 0}^{n - 1} {n - 1 choose j} cdot frac{1}{2n} cdot 1^{frac{2j + 1}{2n} - 1}Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right) \
&= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right)
end{align}
Using the relationship between the Beta and Gamma function:
begin{align}
I_n &= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} Bleft(1 - frac{2j + 1}{2n} , frac{2j + 1}{2n}right) = frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} frac{Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right)}{Gammaleft(1 - frac{2j + 1}{2n} + frac{2j + 1}{2n}right)} \
&= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j}Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right)
end{align}
As $frac{2j + 1}{2n} not in mathbb{Z}$ we can employ Euler's Reflection Formula to yield:
begin{align}
I_n &= frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j}Gammaleft(1 - frac{2j + 1}{2n}right)Gammaleft(frac{2j + 1}{2n}right) = frac{1}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} frac{pi}{sinleft(pi cdot frac{2j + 1}{2n}right)} \&= frac{pi}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} operatorname{cosec}left(frac{pi}{2n}left(2j + 1right) right)
end{align}
Thus,
begin{equation}
int_0^{frac{pi}{2}}frac{1}{sin^{2n}(x) + cos^{2n}(x)}:dx = frac{pi}{2n}sum_{j = 0}^{n - 1} {n - 1 choose j} operatorname{cosec}left(frac{pi}{2n}left(2j + 1right) right)
end{equation}
Edit - Realised that $2n - 1$ should be $n - 1$
Extra Addition:
Using the same method as above an by letting
begin{equation}
S_n(x) = sin^{2n}(x) + cos^{2n}(x)
end{equation}
It becomes rather easy to solve:
begin{equation}
I_{n,m} = int_0^{frac{pi}{2}} frac{S_n(x)}{S_m(x)}:dx
end{equation}
Where $n lt m$. After applying the double angle formulas, $u$-substitution, and half tangent formula we arrive at:
begin{equation}
I_{n,m} = int_0^{infty} frac{S_n(x)}{S_m(x)}:dx = int_0^{infty} frac{t^{2n} + 1}{t^{2m} + 1} left[1 + t^2 right]^{m - n - 1}:dt
end{equation}
Applying the Binomial Expansion:
begin{align}
I_{n,m} &= sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2n} + 1}{t^{2m} + 1} cdot t^{2j}:dt \
&= sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2left(n + jright)} }{t^{2m} + 1} :dt + sum_{j = 0}^{n - 1} {m - n - 1 choose j} int_0^{infty} frac{t^{2j} }{t^{2m} + 1} :dt\
end{align}
Again applying another of my solutions (referenced above) we arrive at:
begin{align}
I_{n,m} &= sum_{j = 0}^{n - 1} {m - n - 1 choose j} frac{pi}{2m}operatorname{cosec}left(frac{pi}{2m}left(2left(n + jright) + 1right) right) + sum_{j = 0}^{n - 1} {m - n - 1 choose j}frac{pi}{2m}operatorname{cosec}left(frac{pi}{2m}left(2j + 1right) right) \
&= frac{pi}{2m}sum_{j = 0}^{n - 1} {m - n - 1 choose j}left[operatorname{cosec}left(frac{pi}{2m}left(2left(n + jright) + 1right) right) + operatorname{cosec}left(frac{pi}{2m}left(2j + 1right) right) right]
end{align}
integration definite-integrals gamma-function trigonometric-integrals beta-function
integration definite-integrals gamma-function trigonometric-integrals beta-function
asked Jan 3 at 2:06
DavidGDavidG
2,047720
asked Jan 3 at 2:06
DavidGDavidG
2,047720
edited Jan 4 at 4:04
DavidG
asked Jan 3 at 2:06
DavidGDavidG
2,047720
asked Jan 3 at 2:06
DavidGDavidG
2,047720
asked Jan 3 at 2:06
DavidGDavidG
2,047720
2,047720
$begingroup$
I think that's the "closest" you can reach to the "closed form"
$endgroup$
– Digamma
Jan 3 at 3:54
$begingroup$
@Digamma - Thanks. Let me know if you think of anything :-)
$endgroup$
– DavidG
Jan 3 at 8:59
$begingroup$
Wikipedia defines "a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations." So if $n$ is finite, you have a closed form. Bravo David, this one is really spectacular.
$endgroup$
– clathratus
Jan 3 at 23:29
$begingroup$
Thanks @clathratus!.
$endgroup$
– DavidG
Jan 4 at 1:18
add a comment |
$begingroup$
I think that's the "closest" you can reach to the "closed form"
$endgroup$
– Digamma
Jan 3 at 3:54
$begingroup$
@Digamma - Thanks. Let me know if you think of anything :-)
$endgroup$
– DavidG
Jan 3 at 8:59
$begingroup$
Wikipedia defines "a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations." So if $n$ is finite, you have a closed form. Bravo David, this one is really spectacular.
$endgroup$
– clathratus
Jan 3 at 23:29
$begingroup$
Thanks @clathratus!.
$endgroup$
– DavidG
Jan 4 at 1:18
$begingroup$
I think that's the "closest" you can reach to the "closed form"
$endgroup$
– Digamma
Jan 3 at 3:54
$begingroup$
I think that's the "closest" you can reach to the "closed form"
$endgroup$
– Digamma
Jan 3 at 3:54
$begingroup$
@Digamma - Thanks. Let me know if you think of anything :-)
$endgroup$
– DavidG
Jan 3 at 8:59
$begingroup$
@Digamma - Thanks. Let me know if you think of anything :-)
$endgroup$
– DavidG
Jan 3 at 8:59
$begingroup$
Wikipedia defines "a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations." So if $n$ is finite, you have a closed form. Bravo David, this one is really spectacular.
$endgroup$
– clathratus
Jan 3 at 23:29
$begingroup$
Wikipedia defines "a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations." So if $n$ is finite, you have a closed form. Bravo David, this one is really spectacular.
$endgroup$
– clathratus
Jan 3 at 23:29
$begingroup$
Thanks @clathratus!.
$endgroup$
– DavidG
Jan 4 at 1:18
$begingroup$
Thanks @clathratus!.
$endgroup$
– DavidG
Jan 4 at 1:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I cannot seem to add much here other than a slightly quicker way for you to reach the integral you arrive at after making a tangent half-angle substitution.
Factoring out a $cos^{2n} x$ term in the denominator of the original integral for $I_n$ we have
$$I_n = int_0^{frac{pi}{2}} frac{sec^{2n} x}{1 + tan^{2n} x} , dx = int_0^{frac{pi}{2}} frac{sec^{2n - 2} x}{1 + tan^{2n} x} sec^2 x , dx.$$
On setting $t = tan x, dt = sec^2 x , dx$, we have
$$I_n = int_0^infty frac{(1 + t^2)^{n - 1}}{1 + t^{2n}} , dt.$$
answered Jan 3 at 4:27
omegadotomegadot
4,9322727
$endgroup$
$begingroup$
Thanks @omegadot - Yes a much quicker way than the approach I took.
$endgroup$
– DavidG
Jan 3 at 4:54
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060193%2fsolving-int-0-frac-pi2-frac1-sin2nx-cos2nx-dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I cannot seem to add much here other than a slightly quicker way for you to reach the integral you arrive at after making a tangent half-angle substitution.
Factoring out a $cos^{2n} x$ term in the denominator of the original integral for $I_n$ we have
$$I_n = int_0^{frac{pi}{2}} frac{sec^{2n} x}{1 + tan^{2n} x} , dx = int_0^{frac{pi}{2}} frac{sec^{2n - 2} x}{1 + tan^{2n} x} sec^2 x , dx.$$
On setting $t = tan x, dt = sec^2 x , dx$, we have
$$I_n = int_0^infty frac{(1 + t^2)^{n - 1}}{1 + t^{2n}} , dt.$$
answered Jan 3 at 4:27
omegadotomegadot
4,9322727
$endgroup$
$begingroup$
Thanks @omegadot - Yes a much quicker way than the approach I took.
$endgroup$
– DavidG
Jan 3 at 4:54
add a comment |
$begingroup$
I cannot seem to add much here other than a slightly quicker way for you to reach the integral you arrive at after making a tangent half-angle substitution.
Factoring out a $cos^{2n} x$ term in the denominator of the original integral for $I_n$ we have
$$I_n = int_0^{frac{pi}{2}} frac{sec^{2n} x}{1 + tan^{2n} x} , dx = int_0^{frac{pi}{2}} frac{sec^{2n - 2} x}{1 + tan^{2n} x} sec^2 x , dx.$$
On setting $t = tan x, dt = sec^2 x , dx$, we have
$$I_n = int_0^infty frac{(1 + t^2)^{n - 1}}{1 + t^{2n}} , dt.$$
answered Jan 3 at 4:27
omegadotomegadot
4,9322727
$endgroup$
$begingroup$
Thanks @omegadot - Yes a much quicker way than the approach I took.
$endgroup$
– DavidG
Jan 3 at 4:54
add a comment |
$begingroup$
I cannot seem to add much here other than a slightly quicker way for you to reach the integral you arrive at after making a tangent half-angle substitution.
Factoring out a $cos^{2n} x$ term in the denominator of the original integral for $I_n$ we have
$$I_n = int_0^{frac{pi}{2}} frac{sec^{2n} x}{1 + tan^{2n} x} , dx = int_0^{frac{pi}{2}} frac{sec^{2n - 2} x}{1 + tan^{2n} x} sec^2 x , dx.$$
On setting $t = tan x, dt = sec^2 x , dx$, we have
$$I_n = int_0^infty frac{(1 + t^2)^{n - 1}}{1 + t^{2n}} , dt.$$
answered Jan 3 at 4:27
omegadotomegadot
4,9322727
$endgroup$
I cannot seem to add much here other than a slightly quicker way for you to reach the integral you arrive at after making a tangent half-angle substitution.
Factoring out a $cos^{2n} x$ term in the denominator of the original integral for $I_n$ we have
$$I_n = int_0^{frac{pi}{2}} frac{sec^{2n} x}{1 + tan^{2n} x} , dx = int_0^{frac{pi}{2}} frac{sec^{2n - 2} x}{1 + tan^{2n} x} sec^2 x , dx.$$
On setting $t = tan x, dt = sec^2 x , dx$, we have
$$I_n = int_0^infty frac{(1 + t^2)^{n - 1}}{1 + t^{2n}} , dt.$$
answered Jan 3 at 4:27
omegadotomegadot
4,9322727
answered Jan 3 at 4:27
omegadotomegadot
4,9322727
answered Jan 3 at 4:27
omegadotomegadot
4,9322727
answered Jan 3 at 4:27
omegadotomegadot
4,9322727
4,9322727
$begingroup$
Thanks @omegadot - Yes a much quicker way than the approach I took.
$endgroup$
– DavidG
Jan 3 at 4:54
add a comment |
$begingroup$
Thanks @omegadot - Yes a much quicker way than the approach I took.
$endgroup$
– DavidG
Jan 3 at 4:54
$begingroup$
Thanks @omegadot - Yes a much quicker way than the approach I took.
$endgroup$
– DavidG
Jan 3 at 4:54
$begingroup$
Thanks @omegadot - Yes a much quicker way than the approach I took.
$endgroup$
– DavidG
Jan 3 at 4:54
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060193%2fsolving-int-0-frac-pi2-frac1-sin2nx-cos2nx-dx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I think that's the "closest" you can reach to the "closed form"
$endgroup$
– Digamma
Jan 3 at 3:54
$begingroup$
@Digamma - Thanks. Let me know if you think of anything :-)
$endgroup$
– DavidG
Jan 3 at 8:59
$begingroup$
Wikipedia defines "a closed-form expression is a mathematical expression that can be evaluated in a finite number of operations." So if $n$ is finite, you have a closed form. Bravo David, this one is really spectacular.
$endgroup$
– clathratus
Jan 3 at 23:29
$begingroup$
Thanks @clathratus!.
$endgroup$
– DavidG
Jan 4 at 1:18