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
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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
$begingroup$
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
asked 2 days ago
JasmineJasmine
393213
$endgroup$
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
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
asked 2 days ago
JasmineJasmine
393213
$endgroup$
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
definite-integrals
asked 2 days ago
JasmineJasmine
393213
asked 2 days ago
JasmineJasmine
393213
edited 2 days ago
Jasmine
asked 2 days ago
JasmineJasmine
393213
asked 2 days ago
JasmineJasmine
393213
asked 2 days ago
JasmineJasmine
393213
393213
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 |
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
12
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
$begingroup$
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
$begingroup$
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
$begingroup$
@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 |
1 Answer
1
active
oldest
votes
$begingroup$
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.
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 |
Your Answer
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1 Answer
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active
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1 Answer
1
active
oldest
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active
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active
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votes
$begingroup$
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.
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.
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.
answered 2 days ago
NewBornMATHNewBornMATH
521111
$endgroup$
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.
answered 2 days ago
NewBornMATHNewBornMATH
521111
edited 2 days ago
answered 2 days ago
NewBornMATHNewBornMATH
521111
answered 2 days ago
NewBornMATHNewBornMATH
521111
answered 2 days ago
NewBornMATHNewBornMATH
521111
521111
$begingroup$
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 |
$begingroup$
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
$begingroup$
Please see the edited question.
$endgroup$
– Jasmine
2 days ago
$begingroup$
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
$begingroup$
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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$begingroup$
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