Why does $int_0^infty cos x^2 dx$ converge? The Next CEO of Stack Overflow$iint_D cos left(...
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Why does $int_0^infty cos x^2 dx$ converge?
The Next CEO of Stack Overflow$iint_D cos left( frac{x-y}{x+y} right),dA$Comparison test for series $sum_{n=1}^{infty}frac{n}{n^3 - 2n + 1}$Simple question about Integral from 0 to f(x) instead of xConvergent subsequences of $x_n = sin n$ and $y_n = cos n$…How to integrate $e^{sin x}(x cos x - tan x sec x)$Does the sum $sum_{n=1}^{infty}{a_nb_n}$ converge(fourier series coefficients)?Discuss the convergence of $int_0^infty x sin e^x , dx$Find $int_0^infty frac{sin(4x)}{x}$Limit of $lim_{t to infty} frac{ int_0^infty cos(x t) e^{-x^k}dx}{int_0^infty cos(x t) e^{-x^p}dx}$Prove that $int_0^{infty} frac{sin x}{x^p}, dx$ converges for $0<p<2$
$begingroup$
I'm a B.C. student. I took a test and a question was:
does $int_0^infty cos x^2 dx$ converge?
- Yes
- No
I undoubtedly answered No because this is going to be somehow periodically, not exactly periodically but after any given number, there is a greater number $s$ that $int_0^s cos x^2 dx = 0$ and a little bit after $s$ the value is not zero. because it is $cos$ you know!
But after the test, I found out that the right answer is Yes and I was wrong.
Now I'm thinking for hours and I really really can't understand it. please help.
integration convergence
asked Mar 16 at 18:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
add a comment |
$begingroup$
I'm a B.C. student. I took a test and a question was:
does $int_0^infty cos x^2 dx$ converge?
- Yes
- No
I undoubtedly answered No because this is going to be somehow periodically, not exactly periodically but after any given number, there is a greater number $s$ that $int_0^s cos x^2 dx = 0$ and a little bit after $s$ the value is not zero. because it is $cos$ you know!
But after the test, I found out that the right answer is Yes and I was wrong.
Now I'm thinking for hours and I really really can't understand it. please help.
integration convergence
asked Mar 16 at 18:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
1
$begingroup$
related: en.wikipedia.org/wiki/Fresnel_integral#Properties
$endgroup$
– Math-fun
Mar 16 at 18:56
add a comment |
$begingroup$
I'm a B.C. student. I took a test and a question was:
does $int_0^infty cos x^2 dx$ converge?
- Yes
- No
I undoubtedly answered No because this is going to be somehow periodically, not exactly periodically but after any given number, there is a greater number $s$ that $int_0^s cos x^2 dx = 0$ and a little bit after $s$ the value is not zero. because it is $cos$ you know!
But after the test, I found out that the right answer is Yes and I was wrong.
Now I'm thinking for hours and I really really can't understand it. please help.
integration convergence
asked Mar 16 at 18:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
$endgroup$
I'm a B.C. student. I took a test and a question was:
does $int_0^infty cos x^2 dx$ converge?
- Yes
- No
I undoubtedly answered No because this is going to be somehow periodically, not exactly periodically but after any given number, there is a greater number $s$ that $int_0^s cos x^2 dx = 0$ and a little bit after $s$ the value is not zero. because it is $cos$ you know!
But after the test, I found out that the right answer is Yes and I was wrong.
Now I'm thinking for hours and I really really can't understand it. please help.
integration convergence
integration convergence
asked Mar 16 at 18:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Mar 16 at 18:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Mar 16 at 18:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Mar 16 at 18:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
asked Mar 16 at 18:33
Peyman mohseni kiasariPeyman mohseni kiasari
13711
13711
1
$begingroup$
related: en.wikipedia.org/wiki/Fresnel_integral#Properties
$endgroup$
– Math-fun
Mar 16 at 18:56
add a comment |
1
$begingroup$
related: en.wikipedia.org/wiki/Fresnel_integral#Properties
$endgroup$
– Math-fun
Mar 16 at 18:56
1
1
$begingroup$
related: en.wikipedia.org/wiki/Fresnel_integral#Properties
$endgroup$
– Math-fun
Mar 16 at 18:56
$begingroup$
related: en.wikipedia.org/wiki/Fresnel_integral#Properties
$endgroup$
– Math-fun
Mar 16 at 18:56
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
To use some elementary methods to prove this integral converges, consider that $$int_0^infty cos(x^2),dx=int_0^1cos(x^2),dx+int_1^inftycos(x^2),dx.$$ The first integral obviously converges since $|cos(x^2)|leq1$. For the second integral, we have $$int_1^inftyfrac{1}{x},xcos(x^2),dx=frac{sin(x^2)}{2x}bigg|_{x=1}^{x=infty}+int_1^inftyfrac{sin(x^2)}{2x^2},dx.$$ This equality follows from integration by parts; the first term above is obviously convergent, while the integral can be bounded above, since $$left|frac{sin(x^2)}{2x^2}right|leqfrac{1}{2x^2},$$and $int x^{-2},dx$ converges as can be shown directly using the power rule for integration.
answered Mar 16 at 18:46
ClaytonClayton
19.5k33288
$endgroup$
$begingroup$
and finally, it is also good to mention that it converges to $sqrt{frac{pi}{8}}$ thanks to en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:00
1
$begingroup$
@Peymanmohsenikiasari: Indeed! To determine the exact value the integral converges to resorts to using complex analysis (contour integration). I thought this might be a bit above a BC calculus class, so I didn't use it and instead chose to use elementary methods. There might be elementary methods to prove the exact value, but I'm not aware of any off the top of my head.
$endgroup$
– Clayton
Mar 16 at 19:03
$begingroup$
your answer was great and so helpful, thank you. I just mention that for "more information" for whoever visit this page in the future.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:07
add a comment |
$begingroup$
Don't be deceived; although $cos x^2$ oscillates, it does at a frequency which increases as $x,$ which makes the argument go to $+infty$ as $x$ becomes arbitrarily large. Therefore, although it oscillates, it does so in such a way that it adds and subtracts sufficiently smaller and smaller areas, and does so faster than $x$ is running away, in order to converge.
I have given a heuristic explanation in the spirit of OP
answered Mar 16 at 18:48
AllawonderAllawonder
1
$endgroup$
1
$begingroup$
Surely the frequency scales as $x$, not $x^2$, since it takes a $O(1/x)$ change in a large $x$ to increase $x^2$ by $2pi$.
$endgroup$
– J.G.
Mar 16 at 18:51
$begingroup$
thanks, nice explanation! what a tricky and beautiful convergence problem it was.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:52
$begingroup$
@J.G. Sure, that's true. Thanks, I got carried away by the extremely loose way I was talking.
$endgroup$
– Allawonder
Mar 16 at 18:58
add a comment |
$begingroup$
The usual idea of a function decaying sufficiently fast for convergence of the improper integral does not apply to sign-changing functions. The reason for the convergence in this case is the highly oscillatory character of the function $cos x^2$ at infinity. The positive and negative "areas" cancel each other and the integral converges.
answered Mar 16 at 18:39
GReyesGReyes
2,32315
$endgroup$
$begingroup$
I don't think that we can say "they cancel each other" that easily. why in cos x^2 "the areas cancel each other" but not in cos ln x? I think this needs better math proof.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:46
$begingroup$
I agree. That was just a rough explanation to make convergence plausible. To justify the claim, one should apply Dirichlet's or Abel's criteria, or a nice elementary argument like Clayton's.
$endgroup$
– GReyes
Mar 16 at 20:13
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
To use some elementary methods to prove this integral converges, consider that $$int_0^infty cos(x^2),dx=int_0^1cos(x^2),dx+int_1^inftycos(x^2),dx.$$ The first integral obviously converges since $|cos(x^2)|leq1$. For the second integral, we have $$int_1^inftyfrac{1}{x},xcos(x^2),dx=frac{sin(x^2)}{2x}bigg|_{x=1}^{x=infty}+int_1^inftyfrac{sin(x^2)}{2x^2},dx.$$ This equality follows from integration by parts; the first term above is obviously convergent, while the integral can be bounded above, since $$left|frac{sin(x^2)}{2x^2}right|leqfrac{1}{2x^2},$$and $int x^{-2},dx$ converges as can be shown directly using the power rule for integration.
answered Mar 16 at 18:46
ClaytonClayton
19.5k33288
$endgroup$
$begingroup$
and finally, it is also good to mention that it converges to $sqrt{frac{pi}{8}}$ thanks to en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:00
1
$begingroup$
@Peymanmohsenikiasari: Indeed! To determine the exact value the integral converges to resorts to using complex analysis (contour integration). I thought this might be a bit above a BC calculus class, so I didn't use it and instead chose to use elementary methods. There might be elementary methods to prove the exact value, but I'm not aware of any off the top of my head.
$endgroup$
– Clayton
Mar 16 at 19:03
$begingroup$
your answer was great and so helpful, thank you. I just mention that for "more information" for whoever visit this page in the future.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:07
add a comment |
$begingroup$
To use some elementary methods to prove this integral converges, consider that $$int_0^infty cos(x^2),dx=int_0^1cos(x^2),dx+int_1^inftycos(x^2),dx.$$ The first integral obviously converges since $|cos(x^2)|leq1$. For the second integral, we have $$int_1^inftyfrac{1}{x},xcos(x^2),dx=frac{sin(x^2)}{2x}bigg|_{x=1}^{x=infty}+int_1^inftyfrac{sin(x^2)}{2x^2},dx.$$ This equality follows from integration by parts; the first term above is obviously convergent, while the integral can be bounded above, since $$left|frac{sin(x^2)}{2x^2}right|leqfrac{1}{2x^2},$$and $int x^{-2},dx$ converges as can be shown directly using the power rule for integration.
answered Mar 16 at 18:46
ClaytonClayton
19.5k33288
$endgroup$
$begingroup$
and finally, it is also good to mention that it converges to $sqrt{frac{pi}{8}}$ thanks to en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:00
1
$begingroup$
@Peymanmohsenikiasari: Indeed! To determine the exact value the integral converges to resorts to using complex analysis (contour integration). I thought this might be a bit above a BC calculus class, so I didn't use it and instead chose to use elementary methods. There might be elementary methods to prove the exact value, but I'm not aware of any off the top of my head.
$endgroup$
– Clayton
Mar 16 at 19:03
$begingroup$
your answer was great and so helpful, thank you. I just mention that for "more information" for whoever visit this page in the future.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:07
add a comment |
$begingroup$
To use some elementary methods to prove this integral converges, consider that $$int_0^infty cos(x^2),dx=int_0^1cos(x^2),dx+int_1^inftycos(x^2),dx.$$ The first integral obviously converges since $|cos(x^2)|leq1$. For the second integral, we have $$int_1^inftyfrac{1}{x},xcos(x^2),dx=frac{sin(x^2)}{2x}bigg|_{x=1}^{x=infty}+int_1^inftyfrac{sin(x^2)}{2x^2},dx.$$ This equality follows from integration by parts; the first term above is obviously convergent, while the integral can be bounded above, since $$left|frac{sin(x^2)}{2x^2}right|leqfrac{1}{2x^2},$$and $int x^{-2},dx$ converges as can be shown directly using the power rule for integration.
answered Mar 16 at 18:46
ClaytonClayton
19.5k33288
$endgroup$
To use some elementary methods to prove this integral converges, consider that $$int_0^infty cos(x^2),dx=int_0^1cos(x^2),dx+int_1^inftycos(x^2),dx.$$ The first integral obviously converges since $|cos(x^2)|leq1$. For the second integral, we have $$int_1^inftyfrac{1}{x},xcos(x^2),dx=frac{sin(x^2)}{2x}bigg|_{x=1}^{x=infty}+int_1^inftyfrac{sin(x^2)}{2x^2},dx.$$ This equality follows from integration by parts; the first term above is obviously convergent, while the integral can be bounded above, since $$left|frac{sin(x^2)}{2x^2}right|leqfrac{1}{2x^2},$$and $int x^{-2},dx$ converges as can be shown directly using the power rule for integration.
answered Mar 16 at 18:46
ClaytonClayton
19.5k33288
answered Mar 16 at 18:46
ClaytonClayton
19.5k33288
answered Mar 16 at 18:46
ClaytonClayton
19.5k33288
answered Mar 16 at 18:46
ClaytonClayton
19.5k33288
19.5k33288
$begingroup$
and finally, it is also good to mention that it converges to $sqrt{frac{pi}{8}}$ thanks to en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:00
1
$begingroup$
@Peymanmohsenikiasari: Indeed! To determine the exact value the integral converges to resorts to using complex analysis (contour integration). I thought this might be a bit above a BC calculus class, so I didn't use it and instead chose to use elementary methods. There might be elementary methods to prove the exact value, but I'm not aware of any off the top of my head.
$endgroup$
– Clayton
Mar 16 at 19:03
$begingroup$
your answer was great and so helpful, thank you. I just mention that for "more information" for whoever visit this page in the future.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:07
add a comment |
$begingroup$
and finally, it is also good to mention that it converges to $sqrt{frac{pi}{8}}$ thanks to en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:00
1
$begingroup$
@Peymanmohsenikiasari: Indeed! To determine the exact value the integral converges to resorts to using complex analysis (contour integration). I thought this might be a bit above a BC calculus class, so I didn't use it and instead chose to use elementary methods. There might be elementary methods to prove the exact value, but I'm not aware of any off the top of my head.
$endgroup$
– Clayton
Mar 16 at 19:03
$begingroup$
your answer was great and so helpful, thank you. I just mention that for "more information" for whoever visit this page in the future.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:07
$begingroup$
and finally, it is also good to mention that it converges to $sqrt{frac{pi}{8}}$ thanks to en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:00
$begingroup$
and finally, it is also good to mention that it converges to $sqrt{frac{pi}{8}}$ thanks to en.wikipedia.org/wiki/Fresnel_integral
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:00
1
1
$begingroup$
@Peymanmohsenikiasari: Indeed! To determine the exact value the integral converges to resorts to using complex analysis (contour integration). I thought this might be a bit above a BC calculus class, so I didn't use it and instead chose to use elementary methods. There might be elementary methods to prove the exact value, but I'm not aware of any off the top of my head.
$endgroup$
– Clayton
Mar 16 at 19:03
$begingroup$
@Peymanmohsenikiasari: Indeed! To determine the exact value the integral converges to resorts to using complex analysis (contour integration). I thought this might be a bit above a BC calculus class, so I didn't use it and instead chose to use elementary methods. There might be elementary methods to prove the exact value, but I'm not aware of any off the top of my head.
$endgroup$
– Clayton
Mar 16 at 19:03
$begingroup$
your answer was great and so helpful, thank you. I just mention that for "more information" for whoever visit this page in the future.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:07
$begingroup$
your answer was great and so helpful, thank you. I just mention that for "more information" for whoever visit this page in the future.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 19:07
add a comment |
$begingroup$
Don't be deceived; although $cos x^2$ oscillates, it does at a frequency which increases as $x,$ which makes the argument go to $+infty$ as $x$ becomes arbitrarily large. Therefore, although it oscillates, it does so in such a way that it adds and subtracts sufficiently smaller and smaller areas, and does so faster than $x$ is running away, in order to converge.
I have given a heuristic explanation in the spirit of OP
answered Mar 16 at 18:48
AllawonderAllawonder
1
$endgroup$
1
$begingroup$
Surely the frequency scales as $x$, not $x^2$, since it takes a $O(1/x)$ change in a large $x$ to increase $x^2$ by $2pi$.
$endgroup$
– J.G.
Mar 16 at 18:51
$begingroup$
thanks, nice explanation! what a tricky and beautiful convergence problem it was.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:52
$begingroup$
@J.G. Sure, that's true. Thanks, I got carried away by the extremely loose way I was talking.
$endgroup$
– Allawonder
Mar 16 at 18:58
add a comment |
$begingroup$
Don't be deceived; although $cos x^2$ oscillates, it does at a frequency which increases as $x,$ which makes the argument go to $+infty$ as $x$ becomes arbitrarily large. Therefore, although it oscillates, it does so in such a way that it adds and subtracts sufficiently smaller and smaller areas, and does so faster than $x$ is running away, in order to converge.
I have given a heuristic explanation in the spirit of OP
answered Mar 16 at 18:48
AllawonderAllawonder
1
$endgroup$
1
$begingroup$
Surely the frequency scales as $x$, not $x^2$, since it takes a $O(1/x)$ change in a large $x$ to increase $x^2$ by $2pi$.
$endgroup$
– J.G.
Mar 16 at 18:51
$begingroup$
thanks, nice explanation! what a tricky and beautiful convergence problem it was.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:52
$begingroup$
@J.G. Sure, that's true. Thanks, I got carried away by the extremely loose way I was talking.
$endgroup$
– Allawonder
Mar 16 at 18:58
add a comment |
$begingroup$
Don't be deceived; although $cos x^2$ oscillates, it does at a frequency which increases as $x,$ which makes the argument go to $+infty$ as $x$ becomes arbitrarily large. Therefore, although it oscillates, it does so in such a way that it adds and subtracts sufficiently smaller and smaller areas, and does so faster than $x$ is running away, in order to converge.
I have given a heuristic explanation in the spirit of OP
answered Mar 16 at 18:48
AllawonderAllawonder
1
$endgroup$
Don't be deceived; although $cos x^2$ oscillates, it does at a frequency which increases as $x,$ which makes the argument go to $+infty$ as $x$ becomes arbitrarily large. Therefore, although it oscillates, it does so in such a way that it adds and subtracts sufficiently smaller and smaller areas, and does so faster than $x$ is running away, in order to converge.
I have given a heuristic explanation in the spirit of OP
answered Mar 16 at 18:48
AllawonderAllawonder
1
edited Mar 16 at 19:00
answered Mar 16 at 18:48
AllawonderAllawonder
1
answered Mar 16 at 18:48
AllawonderAllawonder
1
answered Mar 16 at 18:48
AllawonderAllawonder
1
1
1
$begingroup$
Surely the frequency scales as $x$, not $x^2$, since it takes a $O(1/x)$ change in a large $x$ to increase $x^2$ by $2pi$.
$endgroup$
– J.G.
Mar 16 at 18:51
$begingroup$
thanks, nice explanation! what a tricky and beautiful convergence problem it was.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:52
$begingroup$
@J.G. Sure, that's true. Thanks, I got carried away by the extremely loose way I was talking.
$endgroup$
– Allawonder
Mar 16 at 18:58
add a comment |
1
$begingroup$
Surely the frequency scales as $x$, not $x^2$, since it takes a $O(1/x)$ change in a large $x$ to increase $x^2$ by $2pi$.
$endgroup$
– J.G.
Mar 16 at 18:51
$begingroup$
thanks, nice explanation! what a tricky and beautiful convergence problem it was.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:52
$begingroup$
@J.G. Sure, that's true. Thanks, I got carried away by the extremely loose way I was talking.
$endgroup$
– Allawonder
Mar 16 at 18:58
1
1
$begingroup$
Surely the frequency scales as $x$, not $x^2$, since it takes a $O(1/x)$ change in a large $x$ to increase $x^2$ by $2pi$.
$endgroup$
– J.G.
Mar 16 at 18:51
$begingroup$
Surely the frequency scales as $x$, not $x^2$, since it takes a $O(1/x)$ change in a large $x$ to increase $x^2$ by $2pi$.
$endgroup$
– J.G.
Mar 16 at 18:51
$begingroup$
thanks, nice explanation! what a tricky and beautiful convergence problem it was.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:52
$begingroup$
thanks, nice explanation! what a tricky and beautiful convergence problem it was.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:52
$begingroup$
@J.G. Sure, that's true. Thanks, I got carried away by the extremely loose way I was talking.
$endgroup$
– Allawonder
Mar 16 at 18:58
$begingroup$
@J.G. Sure, that's true. Thanks, I got carried away by the extremely loose way I was talking.
$endgroup$
– Allawonder
Mar 16 at 18:58
add a comment |
$begingroup$
The usual idea of a function decaying sufficiently fast for convergence of the improper integral does not apply to sign-changing functions. The reason for the convergence in this case is the highly oscillatory character of the function $cos x^2$ at infinity. The positive and negative "areas" cancel each other and the integral converges.
answered Mar 16 at 18:39
GReyesGReyes
2,32315
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$begingroup$
I don't think that we can say "they cancel each other" that easily. why in cos x^2 "the areas cancel each other" but not in cos ln x? I think this needs better math proof.
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– Peyman mohseni kiasari
Mar 16 at 18:46
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I agree. That was just a rough explanation to make convergence plausible. To justify the claim, one should apply Dirichlet's or Abel's criteria, or a nice elementary argument like Clayton's.
$endgroup$
– GReyes
Mar 16 at 20:13
add a comment |
$begingroup$
The usual idea of a function decaying sufficiently fast for convergence of the improper integral does not apply to sign-changing functions. The reason for the convergence in this case is the highly oscillatory character of the function $cos x^2$ at infinity. The positive and negative "areas" cancel each other and the integral converges.
answered Mar 16 at 18:39
GReyesGReyes
2,32315
$endgroup$
$begingroup$
I don't think that we can say "they cancel each other" that easily. why in cos x^2 "the areas cancel each other" but not in cos ln x? I think this needs better math proof.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:46
$begingroup$
I agree. That was just a rough explanation to make convergence plausible. To justify the claim, one should apply Dirichlet's or Abel's criteria, or a nice elementary argument like Clayton's.
$endgroup$
– GReyes
Mar 16 at 20:13
add a comment |
$begingroup$
The usual idea of a function decaying sufficiently fast for convergence of the improper integral does not apply to sign-changing functions. The reason for the convergence in this case is the highly oscillatory character of the function $cos x^2$ at infinity. The positive and negative "areas" cancel each other and the integral converges.
answered Mar 16 at 18:39
GReyesGReyes
2,32315
$endgroup$
The usual idea of a function decaying sufficiently fast for convergence of the improper integral does not apply to sign-changing functions. The reason for the convergence in this case is the highly oscillatory character of the function $cos x^2$ at infinity. The positive and negative "areas" cancel each other and the integral converges.
answered Mar 16 at 18:39
GReyesGReyes
2,32315
answered Mar 16 at 18:39
GReyesGReyes
2,32315
answered Mar 16 at 18:39
GReyesGReyes
2,32315
answered Mar 16 at 18:39
GReyesGReyes
2,32315
2,32315
$begingroup$
I don't think that we can say "they cancel each other" that easily. why in cos x^2 "the areas cancel each other" but not in cos ln x? I think this needs better math proof.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:46
$begingroup$
I agree. That was just a rough explanation to make convergence plausible. To justify the claim, one should apply Dirichlet's or Abel's criteria, or a nice elementary argument like Clayton's.
$endgroup$
– GReyes
Mar 16 at 20:13
add a comment |
$begingroup$
I don't think that we can say "they cancel each other" that easily. why in cos x^2 "the areas cancel each other" but not in cos ln x? I think this needs better math proof.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:46
$begingroup$
I agree. That was just a rough explanation to make convergence plausible. To justify the claim, one should apply Dirichlet's or Abel's criteria, or a nice elementary argument like Clayton's.
$endgroup$
– GReyes
Mar 16 at 20:13
$begingroup$
I don't think that we can say "they cancel each other" that easily. why in cos x^2 "the areas cancel each other" but not in cos ln x? I think this needs better math proof.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:46
$begingroup$
I don't think that we can say "they cancel each other" that easily. why in cos x^2 "the areas cancel each other" but not in cos ln x? I think this needs better math proof.
$endgroup$
– Peyman mohseni kiasari
Mar 16 at 18:46
$begingroup$
I agree. That was just a rough explanation to make convergence plausible. To justify the claim, one should apply Dirichlet's or Abel's criteria, or a nice elementary argument like Clayton's.
$endgroup$
– GReyes
Mar 16 at 20:13
$begingroup$
I agree. That was just a rough explanation to make convergence plausible. To justify the claim, one should apply Dirichlet's or Abel's criteria, or a nice elementary argument like Clayton's.
$endgroup$
– GReyes
Mar 16 at 20:13
add a comment |
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related: en.wikipedia.org/wiki/Fresnel_integral#Properties
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– Math-fun
Mar 16 at 18:56