Definite integral of $cos(x^2)$
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Looking at a double integral problem, and the inside integral is $int_y^2^4 y cos(x^2), dx$. I know there's an expression for the indefinite integral of $cos(x^2)$, but what do I do with a definite integral? Or is it just the same?
integration
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up vote
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Looking at a double integral problem, and the inside integral is $int_y^2^4 y cos(x^2), dx$. I know there's an expression for the indefinite integral of $cos(x^2)$, but what do I do with a definite integral? Or is it just the same?
integration
3
The obvious answer is change the order of integration.
– Doug M
Jul 25 at 22:01
Ah, thank you, that explains it!
– BMac
Jul 25 at 22:06
2
What was the whole integral?
– Henry Lee
Jul 26 at 16:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Looking at a double integral problem, and the inside integral is $int_y^2^4 y cos(x^2), dx$. I know there's an expression for the indefinite integral of $cos(x^2)$, but what do I do with a definite integral? Or is it just the same?
integration
Looking at a double integral problem, and the inside integral is $int_y^2^4 y cos(x^2), dx$. I know there's an expression for the indefinite integral of $cos(x^2)$, but what do I do with a definite integral? Or is it just the same?
integration
edited Jul 25 at 22:00
Bernard
110k635103
110k635103
asked Jul 25 at 21:59
BMac
24517
24517
3
The obvious answer is change the order of integration.
– Doug M
Jul 25 at 22:01
Ah, thank you, that explains it!
– BMac
Jul 25 at 22:06
2
What was the whole integral?
– Henry Lee
Jul 26 at 16:48
add a comment |Â
3
The obvious answer is change the order of integration.
– Doug M
Jul 25 at 22:01
Ah, thank you, that explains it!
– BMac
Jul 25 at 22:06
2
What was the whole integral?
– Henry Lee
Jul 26 at 16:48
3
3
The obvious answer is change the order of integration.
– Doug M
Jul 25 at 22:01
The obvious answer is change the order of integration.
– Doug M
Jul 25 at 22:01
Ah, thank you, that explains it!
– BMac
Jul 25 at 22:06
Ah, thank you, that explains it!
– BMac
Jul 25 at 22:06
2
2
What was the whole integral?
– Henry Lee
Jul 26 at 16:48
What was the whole integral?
– Henry Lee
Jul 26 at 16:48
add a comment |Â
1 Answer
1
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Well, I think that you're looking at:
$$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$
Using:
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$
We can write:
$$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Well, I think that you're looking at:
$$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$
Using:
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$
We can write:
$$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$
add a comment |Â
up vote
1
down vote
Well, I think that you're looking at:
$$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$
Using:
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$
We can write:
$$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Well, I think that you're looking at:
$$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$
Using:
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$
We can write:
$$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$
Well, I think that you're looking at:
$$mathcalI_spacealphaleft(beta,etaright):=int_alpha^betaint_texty^2^etatextycdotcosleft(x^2right)spacetextdxspacetextdtexty=int_alpha^betatextycdotleftint_texty^2^etacosleft(x^2right)spacetextdxrightspacetextdtextytag1$$
Using:
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textn=cosleft(x^2right)tag2$$
We can write:
$$mathcalI_spacealphaleft(beta,etaright)=int_alpha^betatextycdotleftint_texty^2^etasum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdot x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftint_texty^2^eta x^4textnspacetextdxrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotint_alpha^betatextycdotleftfraceta^1+4textn1+4textn-fractexty^2left(1+4textnright)1+4textnrightspacetextdtexty=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotint_alpha^betatextyspacetextdtexty-frac11+4textncdotint_alpha^betatexty^1+2left(1+4textnright)spacetextdtextyright=$$
$$sum_textn=0^inftyfracleft(-1right)^textnleft(2textnright)!cdotleftfraceta^1+4textn1+4textncdotfracbeta^2-alpha^22-frac11+4textncdotleft(fracbeta^2+8textn2+8textn-fracalpha^2+8textn2+8textnright)righttag3$$
answered Jul 28 at 11:50


Jan
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3
The obvious answer is change the order of integration.
– Doug M
Jul 25 at 22:01
Ah, thank you, that explains it!
– BMac
Jul 25 at 22:06
2
What was the whole integral?
– Henry Lee
Jul 26 at 16:48