Can anyone help in $int_0^pi (sin x )^{cos x} dx$? [closed]
$begingroup$
I am trying to solve the following definite integral;
$int_0^pi (sin x )^{cos x} dx$?
definite-integrals
edited Dec 8 '18 at 14:07
amWhy
1
asked Dec 8 '18 at 13:25
AlbertAlbert
344
$endgroup$
closed as off-topic by Martin R, mrtaurho, metamorphy, Martín Vacas Vignolo, Paul Frost Dec 30 '18 at 14:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, mrtaurho, metamorphy, Martín Vacas Vignolo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I am trying to solve the following definite integral;
$int_0^pi (sin x )^{cos x} dx$?
definite-integrals
edited Dec 8 '18 at 14:07
amWhy
1
asked Dec 8 '18 at 13:25
AlbertAlbert
344
$endgroup$
closed as off-topic by Martin R, mrtaurho, metamorphy, Martín Vacas Vignolo, Paul Frost Dec 30 '18 at 14:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, mrtaurho, metamorphy, Martín Vacas Vignolo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@OmG Your input is wrong. You computed the integral of the function $xcdot sin(x)^{cos(x)}$ instead of $sin(x)^{cos(x)}$. It does not alter the fact that it does not converge but I would say it is important to be precise concerning the integrand.
$endgroup$
– mrtaurho
Dec 8 '18 at 13:51
add a comment |
$begingroup$
I am trying to solve the following definite integral;
$int_0^pi (sin x )^{cos x} dx$?
definite-integrals
edited Dec 8 '18 at 14:07
amWhy
1
asked Dec 8 '18 at 13:25
AlbertAlbert
344
$endgroup$
I am trying to solve the following definite integral;
$int_0^pi (sin x )^{cos x} dx$?
definite-integrals
definite-integrals
edited Dec 8 '18 at 14:07
amWhy
1
asked Dec 8 '18 at 13:25
AlbertAlbert
344
edited Dec 8 '18 at 14:07
amWhy
1
asked Dec 8 '18 at 13:25
AlbertAlbert
344
edited Dec 8 '18 at 14:07
amWhy
1
edited Dec 8 '18 at 14:07
amWhy
1
edited Dec 8 '18 at 14:07
amWhy
1
1
asked Dec 8 '18 at 13:25
AlbertAlbert
344
asked Dec 8 '18 at 13:25
AlbertAlbert
344
asked Dec 8 '18 at 13:25
AlbertAlbert
344
344
closed as off-topic by Martin R, mrtaurho, metamorphy, Martín Vacas Vignolo, Paul Frost Dec 30 '18 at 14:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, mrtaurho, metamorphy, Martín Vacas Vignolo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Martin R, mrtaurho, metamorphy, Martín Vacas Vignolo, Paul Frost Dec 30 '18 at 14:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Martin R, mrtaurho, metamorphy, Martín Vacas Vignolo, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
@OmG Your input is wrong. You computed the integral of the function $xcdot sin(x)^{cos(x)}$ instead of $sin(x)^{cos(x)}$. It does not alter the fact that it does not converge but I would say it is important to be precise concerning the integrand.
$endgroup$
– mrtaurho
Dec 8 '18 at 13:51
add a comment |
$begingroup$
@OmG Your input is wrong. You computed the integral of the function $xcdot sin(x)^{cos(x)}$ instead of $sin(x)^{cos(x)}$. It does not alter the fact that it does not converge but I would say it is important to be precise concerning the integrand.
$endgroup$
– mrtaurho
Dec 8 '18 at 13:51
$begingroup$
@OmG Your input is wrong. You computed the integral of the function $xcdot sin(x)^{cos(x)}$ instead of $sin(x)^{cos(x)}$. It does not alter the fact that it does not converge but I would say it is important to be precise concerning the integrand.
$endgroup$
– mrtaurho
Dec 8 '18 at 13:51
$begingroup$
@OmG Your input is wrong. You computed the integral of the function $xcdot sin(x)^{cos(x)}$ instead of $sin(x)^{cos(x)}$. It does not alter the fact that it does not converge but I would say it is important to be precise concerning the integrand.
$endgroup$
– mrtaurho
Dec 8 '18 at 13:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The integral does not exist, because $sin(pi-epsilon) = epsilon + o(epsilon^3)$ and $cos(pi-epsilon) = -1 + epsilon^2/2 + o(epsilon^3),$ i.e. the integrand has a pole at $x=pi.$
answered Dec 8 '18 at 13:50
gammatestergammatester
16.7k21633
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The integral does not exist, because $sin(pi-epsilon) = epsilon + o(epsilon^3)$ and $cos(pi-epsilon) = -1 + epsilon^2/2 + o(epsilon^3),$ i.e. the integrand has a pole at $x=pi.$
answered Dec 8 '18 at 13:50
gammatestergammatester
16.7k21633
$endgroup$
add a comment |
$begingroup$
The integral does not exist, because $sin(pi-epsilon) = epsilon + o(epsilon^3)$ and $cos(pi-epsilon) = -1 + epsilon^2/2 + o(epsilon^3),$ i.e. the integrand has a pole at $x=pi.$
answered Dec 8 '18 at 13:50
gammatestergammatester
16.7k21633
$endgroup$
add a comment |
$begingroup$
The integral does not exist, because $sin(pi-epsilon) = epsilon + o(epsilon^3)$ and $cos(pi-epsilon) = -1 + epsilon^2/2 + o(epsilon^3),$ i.e. the integrand has a pole at $x=pi.$
answered Dec 8 '18 at 13:50
gammatestergammatester
16.7k21633
$endgroup$
The integral does not exist, because $sin(pi-epsilon) = epsilon + o(epsilon^3)$ and $cos(pi-epsilon) = -1 + epsilon^2/2 + o(epsilon^3),$ i.e. the integrand has a pole at $x=pi.$
answered Dec 8 '18 at 13:50
gammatestergammatester
16.7k21633
edited Dec 8 '18 at 13:59
answered Dec 8 '18 at 13:50
gammatestergammatester
16.7k21633
answered Dec 8 '18 at 13:50
gammatestergammatester
16.7k21633
answered Dec 8 '18 at 13:50
gammatestergammatester
16.7k21633
16.7k21633
add a comment |
add a comment |
$begingroup$
@OmG Your input is wrong. You computed the integral of the function $xcdot sin(x)^{cos(x)}$ instead of $sin(x)^{cos(x)}$. It does not alter the fact that it does not converge but I would say it is important to be precise concerning the integrand.
$endgroup$
– mrtaurho
Dec 8 '18 at 13:51