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Finding the reason behind the value of the integral.

Not sure how to solve this integralDoes adding an odd function to the numerator of an even function integrand, keep the value of the integral (over $(-infty,infty)$) same?Indefinite integral of absolute valueA difficult trigonometric integral involving absolute valueHelp with Differentiating through the Integral SignDefinite integration using even and odd functionsFinding the value of a definite integralApproximation of integral of error function?Integral of a function not defined at all pointsProve that the integral $int_0^inftybigg|fracsin(pi x)xbigg|dx$ diverges

16

9

$begingroup$

I was just trying to find $$int_0^pi / 2fracsin9xsinx,dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.

I can't figure out the reason and would like to know why this is happening.

---

Edit: It can also be noted that $$int_api^bpi fracsin9xsinx,dx =(b-a)pi$$ where $a,b$ are integers.

definite-integrals

share|cite|improve this question

edited 2 days ago

Jasmine

asked 2 days ago

JasmineJasmine

393213

$endgroup$

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  $begingroup$
  The following identity seems like it may help:$$fracsin((n+1/2)thetasin(theta/2)=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
  $endgroup$
  – Semiclassical
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the statement about the integral from $api$ to $bpi$ is incorrect. Setting $a=frac18$ and $b=frac16,$ Wolfram Alpha says the integral is a negative number, not $(frac16-frac18)pi.$ Did you mean to say instead that $a$ and $b$ are integers?
  $endgroup$
  – David K
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @David I checked for few other rational a and b and they satisfied but you are right.
  $endgroup$
  – Jasmine
  2 days ago
  

  
  
  
  

add a comment | 

16

9

$begingroup$

I was just trying to find $$int_0^pi / 2fracsin9xsinx,dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.

I can't figure out the reason and would like to know why this is happening.

---

Edit: It can also be noted that $$int_api^bpi fracsin9xsinx,dx =(b-a)pi$$ where $a,b$ are integers.

definite-integrals

share|cite|improve this question

edited 2 days ago

Jasmine

asked 2 days ago

JasmineJasmine

393213

$endgroup$

• 
  
  
  

  
  12
  

  
  
  
  
  
  

  
  $begingroup$
  The following identity seems like it may help:$$fracsin((n+1/2)thetasin(theta/2)=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
  $endgroup$
  – Semiclassical
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the statement about the integral from $api$ to $bpi$ is incorrect. Setting $a=frac18$ and $b=frac16,$ Wolfram Alpha says the integral is a negative number, not $(frac16-frac18)pi.$ Did you mean to say instead that $a$ and $b$ are integers?
  $endgroup$
  – David K
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @David I checked for few other rational a and b and they satisfied but you are right.
  $endgroup$
  – Jasmine
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

I was just trying to find $$int_0^pi / 2fracsin9xsinx,dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.

I can't figure out the reason and would like to know why this is happening.

---

Edit: It can also be noted that $$int_api^bpi fracsin9xsinx,dx =(b-a)pi$$ where $a,b$ are integers.

definite-integrals

share|cite|improve this question

edited 2 days ago

Jasmine

asked 2 days ago

JasmineJasmine

393213

$endgroup$

share|cite|improve this question

edited 2 days ago

Jasmine

asked 2 days ago

JasmineJasmine

393213

share|cite|improve this question

edited 2 days ago

Jasmine

asked 2 days ago

JasmineJasmine

393213

asked 2 days ago

JasmineJasmine

393213

asked 2 days ago

JasmineJasmine

393213

1 Answer
1

active

oldest

votes

14

$begingroup$

Hint

Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$

$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.

Edit
(As OP has changed the question a bit)

Now consider$I(n)=int_api^bpi fracsin(nx)sin x dx$

From(1) $$implies 2int_api^bpi cos(2mx)dx=2bigg[fracsin(2mx)2mbigg]_api^bpi$$
$$implies I(2m+1)-I(2m-1)=frac1n bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided $a,b in mathbbZ $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$

Hence

$$ bbox[5px,border:2px solid blue]
int_api^bpi fracsin(nx)sin x dx=(b-a)pi

$$
When n is odd.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

NewBornMATHNewBornMATH

521111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please see the edited question.
  $endgroup$
  – Jasmine
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope it helps else Let me know how i can improve my answer.
  $endgroup$
  – NewBornMATH
  2 days ago
  
  

  
  
  
  

add a comment | 

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14

$begingroup$

Hint

Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$

$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.

Edit
(As OP has changed the question a bit)

Now consider$I(n)=int_api^bpi fracsin(nx)sin x dx$

From(1) $$implies 2int_api^bpi cos(2mx)dx=2bigg[fracsin(2mx)2mbigg]_api^bpi$$
$$implies I(2m+1)-I(2m-1)=frac1n bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided $a,b in mathbbZ $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$

Hence

$$ bbox[5px,border:2px solid blue]
int_api^bpi fracsin(nx)sin x dx=(b-a)pi

$$
When n is odd.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

NewBornMATHNewBornMATH

521111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please see the edited question.
  $endgroup$
  – Jasmine
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope it helps else Let me know how i can improve my answer.
  $endgroup$
  – NewBornMATH
  2 days ago
  
  

  
  
  
  

add a comment | 

14

$begingroup$

Hint

Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$

$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.

Edit
(As OP has changed the question a bit)

Now consider$I(n)=int_api^bpi fracsin(nx)sin x dx$

From(1) $$implies 2int_api^bpi cos(2mx)dx=2bigg[fracsin(2mx)2mbigg]_api^bpi$$
$$implies I(2m+1)-I(2m-1)=frac1n bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided $a,b in mathbbZ $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$

Hence

$$ bbox[5px,border:2px solid blue]
int_api^bpi fracsin(nx)sin x dx=(b-a)pi

$$
When n is odd.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

NewBornMATHNewBornMATH

521111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please see the edited question.
  $endgroup$
  – Jasmine
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hope it helps else Let me know how i can improve my answer.
  $endgroup$
  – NewBornMATH
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint

Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$

$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx.......(1)$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.

Edit
(As OP has changed the question a bit)

Now consider$I(n)=int_api^bpi fracsin(nx)sin x dx$

From(1) $$implies 2int_api^bpi cos(2mx)dx=2bigg[fracsin(2mx)2mbigg]_api^bpi$$
$$implies I(2m+1)-I(2m-1)=frac1n bigg[sin(2pi bx)-sin(2pi ax)bigg]=0$$
Provided $a,b in mathbbZ $
$$implies I(2m+1)=I(2m-1)$$
Now As $I(1)=(b-a)pi$

Hence

$$ bbox[5px,border:2px solid blue]
int_api^bpi fracsin(nx)sin x dx=(b-a)pi

$$
When n is odd.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

NewBornMATHNewBornMATH

521111

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

NewBornMATHNewBornMATH

521111

answered 2 days ago

NewBornMATHNewBornMATH

521111

answered 2 days ago

NewBornMATHNewBornMATH

521111