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Trigonometric residue integral

0

1

$begingroup$

The trigonometric complex integral is
$$
int_0^{2pi} frac 1{(2+sin theta)^2} dtheta.
$$
My answer is below, and it is correct because I verified this numerically on WolframAlpha: https://www.wolframalpha.com/input/?i=%5Cint_0%5E(2pi)+1%2F%7B(2%2B%5Csin+%5Ctheta)%5E2%7D+d%5Ctheta

complex-analysis proof-verification complex-numbers residue-calculus

share|cite|improve this question

edited Mar 24 at 20:39

Cookie

asked Mar 24 at 19:49

CookieCookie

8,803123985

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Do you want an alternate proof?
  $endgroup$
  – clathratus
  Mar 24 at 19:52
  

  
  
  
  

add a comment | 

0

1

$begingroup$

The trigonometric complex integral is
$$
int_0^{2pi} frac 1{(2+sin theta)^2} dtheta.
$$
My answer is below, and it is correct because I verified this numerically on WolframAlpha: https://www.wolframalpha.com/input/?i=%5Cint_0%5E(2pi)+1%2F%7B(2%2B%5Csin+%5Ctheta)%5E2%7D+d%5Ctheta

complex-analysis proof-verification complex-numbers residue-calculus

share|cite|improve this question

edited Mar 24 at 20:39

Cookie

asked Mar 24 at 19:49

CookieCookie

8,803123985

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Do you want an alternate proof?
  $endgroup$
  – clathratus
  Mar 24 at 19:52
  

  
  
  
  

add a comment | 

$begingroup$

The trigonometric complex integral is
$$
int_0^{2pi} frac 1{(2+sin theta)^2} dtheta.
$$
My answer is below, and it is correct because I verified this numerically on WolframAlpha: https://www.wolframalpha.com/input/?i=%5Cint_0%5E(2pi)+1%2F%7B(2%2B%5Csin+%5Ctheta)%5E2%7D+d%5Ctheta

complex-analysis proof-verification complex-numbers residue-calculus

share|cite|improve this question

edited Mar 24 at 20:39

Cookie

asked Mar 24 at 19:49

CookieCookie

8,803123985

$endgroup$

share|cite|improve this question

edited Mar 24 at 20:39

Cookie

asked Mar 24 at 19:49

CookieCookie

8,803123985

share|cite|improve this question

edited Mar 24 at 20:39

Cookie

asked Mar 24 at 19:49

CookieCookie

8,803123985

asked Mar 24 at 19:49

CookieCookie

8,803123985

asked Mar 24 at 19:49

CookieCookie

8,803123985

1 Answer
1

active

oldest

votes

1

$begingroup$

I applied the substitution $z=e^{itheta}$ and then used the residue theorem to obtain
begin{align*}
int_0^{2pi} frac 1{(2+sin theta)^2} dtheta &= int_{|z|=1} frac 1{(2+frac 1{2i}(z-frac 1z))^2} frac{dz}{iz} \
&= 4iint_{|z|=1} frac z{(z^2+4iz-1)^2} dz \
&= 4i cdot 2pi i operatorname{Res}_{z=(-2+sqrt 3)i} frac z{(z^2+4iz-1)^2} \
&= -8pi frac d{dz}left.left((z-(-2+sqrt 3)i)^2 frac z{(z^2+4iz-1)^2} right)rightvert_{z=(-2+sqrt 3)i} \
&= -8pi frac d{dz}left.left(frac z{(z-(-2-sqrt 3)i)^2}right)rightvert_{z=(-2+sqrt 3)i} \
&= -8pileft.left(frac{-z^2-(2+sqrt 3)^2}{(z+(2+sqrt 3)i)^4}right)rightvert_{z=(-2+sqrt 3)i} \
&= boxed{frac{4pisqrt 3}9}
end{align*}

share|cite|improve this answer

edited Mar 25 at 1:35

answered Mar 24 at 20:39

CookieCookie

8,803123985

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your proof is good. Not sure what sort of answer you are looking for though
  $endgroup$
  – Brevan Ellefsen
  Mar 25 at 21:25
  

  
  
  
  

add a comment | 

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1

$begingroup$

I applied the substitution $z=e^{itheta}$ and then used the residue theorem to obtain
begin{align*}
int_0^{2pi} frac 1{(2+sin theta)^2} dtheta &= int_{|z|=1} frac 1{(2+frac 1{2i}(z-frac 1z))^2} frac{dz}{iz} \
&= 4iint_{|z|=1} frac z{(z^2+4iz-1)^2} dz \
&= 4i cdot 2pi i operatorname{Res}_{z=(-2+sqrt 3)i} frac z{(z^2+4iz-1)^2} \
&= -8pi frac d{dz}left.left((z-(-2+sqrt 3)i)^2 frac z{(z^2+4iz-1)^2} right)rightvert_{z=(-2+sqrt 3)i} \
&= -8pi frac d{dz}left.left(frac z{(z-(-2-sqrt 3)i)^2}right)rightvert_{z=(-2+sqrt 3)i} \
&= -8pileft.left(frac{-z^2-(2+sqrt 3)^2}{(z+(2+sqrt 3)i)^4}right)rightvert_{z=(-2+sqrt 3)i} \
&= boxed{frac{4pisqrt 3}9}
end{align*}

share|cite|improve this answer

edited Mar 25 at 1:35

answered Mar 24 at 20:39

CookieCookie

8,803123985

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your proof is good. Not sure what sort of answer you are looking for though
  $endgroup$
  – Brevan Ellefsen
  Mar 25 at 21:25
  

  
  
  
  

add a comment | 

1

$begingroup$

I applied the substitution $z=e^{itheta}$ and then used the residue theorem to obtain
begin{align*}
int_0^{2pi} frac 1{(2+sin theta)^2} dtheta &= int_{|z|=1} frac 1{(2+frac 1{2i}(z-frac 1z))^2} frac{dz}{iz} \
&= 4iint_{|z|=1} frac z{(z^2+4iz-1)^2} dz \
&= 4i cdot 2pi i operatorname{Res}_{z=(-2+sqrt 3)i} frac z{(z^2+4iz-1)^2} \
&= -8pi frac d{dz}left.left((z-(-2+sqrt 3)i)^2 frac z{(z^2+4iz-1)^2} right)rightvert_{z=(-2+sqrt 3)i} \
&= -8pi frac d{dz}left.left(frac z{(z-(-2-sqrt 3)i)^2}right)rightvert_{z=(-2+sqrt 3)i} \
&= -8pileft.left(frac{-z^2-(2+sqrt 3)^2}{(z+(2+sqrt 3)i)^4}right)rightvert_{z=(-2+sqrt 3)i} \
&= boxed{frac{4pisqrt 3}9}
end{align*}

share|cite|improve this answer

edited Mar 25 at 1:35

answered Mar 24 at 20:39

CookieCookie

8,803123985

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your proof is good. Not sure what sort of answer you are looking for though
  $endgroup$
  – Brevan Ellefsen
  Mar 25 at 21:25
  

  
  
  
  

add a comment | 

$begingroup$

I applied the substitution $z=e^{itheta}$ and then used the residue theorem to obtain
begin{align*}
int_0^{2pi} frac 1{(2+sin theta)^2} dtheta &= int_{|z|=1} frac 1{(2+frac 1{2i}(z-frac 1z))^2} frac{dz}{iz} \
&= 4iint_{|z|=1} frac z{(z^2+4iz-1)^2} dz \
&= 4i cdot 2pi i operatorname{Res}_{z=(-2+sqrt 3)i} frac z{(z^2+4iz-1)^2} \
&= -8pi frac d{dz}left.left((z-(-2+sqrt 3)i)^2 frac z{(z^2+4iz-1)^2} right)rightvert_{z=(-2+sqrt 3)i} \
&= -8pi frac d{dz}left.left(frac z{(z-(-2-sqrt 3)i)^2}right)rightvert_{z=(-2+sqrt 3)i} \
&= -8pileft.left(frac{-z^2-(2+sqrt 3)^2}{(z+(2+sqrt 3)i)^4}right)rightvert_{z=(-2+sqrt 3)i} \
&= boxed{frac{4pisqrt 3}9}
end{align*}

share|cite|improve this answer

edited Mar 25 at 1:35

answered Mar 24 at 20:39

CookieCookie

8,803123985

$endgroup$

share|cite|improve this answer

edited Mar 25 at 1:35

answered Mar 24 at 20:39

CookieCookie

8,803123985

answered Mar 24 at 20:39

CookieCookie

8,803123985

answered Mar 24 at 20:39

CookieCookie

8,803123985