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David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250 Posted: Sun Jul 09, 2006 12:45 pm    Post subject: Re: Real Integral w/ Complex Analysis On Sun, 09 Jul 2006 01:08:14 EDT, Narcoleptic Insomniac
<i_have_narcoleptic_insomnia@yahoo.com> wrote:

 Quote: On Jul 8, 2006 3:00 PM, David C. Ullrich wrote: On Sat, 08 Jul 2006 11:26:27 EDT, Narcoleptic Insomniac wrote: On Jul 8, 2006 8:16 AM CT, Narcoleptic Insomniac wrote: All the real integrals I've done with complex analysis have made use of residues (which is one of the coolest things I've seen). Anyways, I've been stumped on the following exercise for awhile which doesn't seem to involve residues. The exercise is to evaluate the integral int_{0}^{oo} cos(x^2) dx ...using complex analysis. I'm supposed consider the real part of e^{iz^2} along a particular contour G. Here a picture would be worth 1000 words, but I'll give my parametrization of G: G = g_1 + g_2 - g_3 g_1: [0, 1] -> C g_1(t) = A * t ...where A is a positive real number, g_2: [0, 1] -> C g_2(t) = A * e^{t * i * pi / 4}, g_3: [0, 1] -> C g_3(t) = A * t * e^{pi * i / 4}. A friend suggested to me a better, more natural parametrization g_1: [0, A] -> C g_1(t) = A Huh? Clearly I made a typo and it should be 't' not 'A'. g_3: [0, A] -> C g_3(t) = t * e^{pi * i / 4} Essentially, G is a pie shaped wedge that begins at the origin, travels along the real axis to the positive number A, goes up in a circular arc pi/4 radians, and then returns to the origin. I'm thinking that since e^{iz^2} is entire we have int_G e^{iz^2} dz = 0 ...by Cauchy's theorem so... int_{g_1} e^{iz^2} dz = int_{g_3 - g_2} e^{iz^2} dz. So if I could evaluate the latter, taking the real part and the limit as A -> oo, then I'd get the value for the cos(x^2) integral. However, when I try do this directly with my parametrization I end up with some nasty integrals. With the better parametrization for g_3 I have int_{g_3} e^{i z^2} dz = e^{pi * i / 4} int_{0, A} e^{-t^2} dt ..so taking the real part of this as A -> oo I get... 1 / sqrt(2) * int_{0, oo} e^{-t^2} dt. Using polar coordinates I can show this last integral to be sqrt(pi) / 2 (is there a better way to do this?) so lim_{A -> oo} Re[int_{g_3} e^{i z^2} dz] = 1/2 * (pi/2)^(1/2). Since this should be the answer all I need to do now is show that lim_{A -> oo} Re[int_{-g_2} e^{i z^2} dz] = 0. I was told that the Jordan lemma would help but I don't see how to apply it in this case. That "Jordan lemma" thing is incredibly stupid. Why? Because it's just a special case of a much more powerful result, _and_ the more general result has exactly the same simple proof! Better Lemma: Suppose that f_n:[a,b] -> C is continuous for each n. Suppose that the f_n are uniformly bounded, |f_n(x)| <= M for all n and all x. Suppose that for every d > 0 we have f_n -> 0 uniformly on [a+d, b-d] as n -> infinity. Then int_a^b f_n -> 0. This in turn is of course just a special case of the Dominated Convergence Theorem from real analysis. But DCT is not quite trivial, while the Better Lemma _is_ trivial: Pf: Let eps > 0. Choose d > 0 such that 2dM < eps/2 Choose N so that |f_n(x)| < eps/(2(b-a)) for all n > N and all x in [a+d, b-d]. Then it follows that int_a^b |f_n(x)| < eps for all n > N (split the integral into two pieces in the obvious way and apply one inequality above to each piece.) QED. I dunno if the suggestion was "incredibly stupid", but I do thank you for showing me that the Jordan lemma is just a specific instance of this general result.

Hmm, looking back I see that my statement
'That "Jordan lemma" thing is incredibly stupid.'
was a little ambiguous. I certainly didn't mean to
say that the suggestion that you use it was stupid.
It's Jordan's Lemma itself, and/or the fact that
it's included in books on complex analysis,
that's incredibly stupid.

 Quote: In this particular problem though I was thinking that on g_2 I have z = A e^{pi i t /4} so |e^{i z^2} dz| = e^{-A^2 sin(pi t / 2)} dt. Now since 0 <= t <= 1 here I have -sin(pi t / 2) <= -t / 2, ..which implies... int_{g_2} |e^{i z^2} dz| <= int_{0}^{1} e^{-A^2 t / 2} dt, the latter of which vanishes as A -> oo. Thanks again to you David and Badger for your time. Regards, Kyle Czarnecki I've read that this is a kind of Fresnel integral and I know the answer is 1/2 * (pi / 2)^(1/2), but I'm really getting stuck so I'd be greatful for any help. Thanks In Advance, Kyle Czarnecki ************************ David C. Ullrich

************************

David C. Ullrich G. A. Edgar
science forum Guru

Joined: 29 Apr 2005
Posts: 470 Posted: Sun Jul 09, 2006 12:28 pm    Post subject: Re: Real Integral w/ Complex Analysis J. H. Cadwell, "Three Integrals," Math. Gazette 31 (1947) 239--240 integrals of sin(x^2), cos(x^2), exp(i x^2) from -infinity to infinity done using coutour integrals around a rectangle... "the integrand is not at all evident" ... L. Mirsky, "The probability integral," Math. Gazette 33 (1949) 279 integral of exp(-x^2) from -infinity to infinity done using a contour integral around a paralellogram... "neither the integrand nor the paralellogram is at all evident"... a recent account: D. Desbrow, Amer. Math. Monthly, October 1998, pp. 726--731 -- G. A. Edgar http://www.math.ohio-state.edu/~edgar/ Narcoleptic Insomniac
science forum Guru

Joined: 02 May 2005
Posts: 323 Posted: Sun Jul 09, 2006 5:08 am    Post subject: Re: Real Integral w/ Complex Analysis On Jul 8, 2006 3:00 PM, David C. Ullrich wrote:

 Quote: On Sat, 08 Jul 2006 11:26:27 EDT, Narcoleptic Insomniac wrote: On Jul 8, 2006 8:16 AM CT, Narcoleptic Insomniac wrote: All the real integrals I've done with complex analysis have made use of residues (which is one of the coolest things I've seen). Anyways, I've been stumped on the following exercise for awhile which doesn't seem to involve residues. The exercise is to evaluate the integral int_{0}^{oo} cos(x^2) dx ...using complex analysis. I'm supposed consider the real part of e^{iz^2} along a particular contour G. Here a picture would be worth 1000 words, but I'll give my parametrization of G: G = g_1 + g_2 - g_3 g_1: [0, 1] -> C g_1(t) = A * t ...where A is a positive real number, g_2: [0, 1] -> C g_2(t) = A * e^{t * i * pi / 4}, g_3: [0, 1] -> C g_3(t) = A * t * e^{pi * i / 4}. A friend suggested to me a better, more natural parametrization g_1: [0, A] -> C g_1(t) = A Huh?

Clearly I made a typo and it should be 't' not 'A'.

 Quote: g_3: [0, A] -> C g_3(t) = t * e^{pi * i / 4} Essentially, G is a pie shaped wedge that begins at the origin, travels along the real axis to the positive number A, goes up in a circular arc pi/4 radians, and then returns to the origin. I'm thinking that since e^{iz^2} is entire we have int_G e^{iz^2} dz = 0 ...by Cauchy's theorem so... int_{g_1} e^{iz^2} dz = int_{g_3 - g_2} e^{iz^2} dz. So if I could evaluate the latter, taking the real part and the limit as A -> oo, then I'd get the value for the cos(x^2) integral. However, when I try do this directly with my parametrization I end up with some nasty integrals. With the better parametrization for g_3 I have int_{g_3} e^{i z^2} dz = e^{pi * i / 4} int_{0, A} e^{-t^2} dt ..so taking the real part of this as A -> oo I get... 1 / sqrt(2) * int_{0, oo} e^{-t^2} dt. Using polar coordinates I can show this last integral to be sqrt(pi) / 2 (is there a better way to do this?) so lim_{A -> oo} Re[int_{g_3} e^{i z^2} dz] = 1/2 * (pi/2)^(1/2). Since this should be the answer all I need to do now is show that lim_{A -> oo} Re[int_{-g_2} e^{i z^2} dz] = 0. I was told that the Jordan lemma would help but I don't see how to apply it in this case. That "Jordan lemma" thing is incredibly stupid. Why? Because it's just a special case of a much more powerful result, _and_ the more general result has exactly the same simple proof! Better Lemma: Suppose that f_n:[a,b] -> C is continuous for each n. Suppose that the f_n are uniformly bounded, |f_n(x)| <= M for all n and all x. Suppose that for every d > 0 we have f_n -> 0 uniformly on [a+d, b-d] as n -> infinity. Then int_a^b f_n -> 0. This in turn is of course just a special case of the Dominated Convergence Theorem from real analysis. But DCT is not quite trivial, while the Better Lemma _is_ trivial: Pf: Let eps > 0. Choose d > 0 such that 2dM < eps/2 Choose N so that |f_n(x)| < eps/(2(b-a)) for all n > N and all x in [a+d, b-d]. Then it follows that int_a^b |f_n(x)| < eps for all n > N (split the integral into two pieces in the obvious way and apply one inequality above to each piece.) QED.

I dunno if the suggestion was "incredibly stupid", but I
do thank you for showing me that the Jordan lemma is just
a specific instance of this general result.

In this particular problem though I was thinking that on
g_2 I have z = A e^{pi i t /4} so

|e^{i z^2} dz| = e^{-A^2 sin(pi t / 2)} dt.

Now since 0 <= t <= 1 here I have

-sin(pi t / 2) <= -t / 2,

...which implies...

int_{g_2} |e^{i z^2} dz| <= int_{0}^{1} e^{-A^2 t / 2} dt,

the latter of which vanishes as A -> oo.

Thanks again to you David and Badger for your time.

Regards,
Kyle Czarnecki

 Quote: I've read that this is a kind of Fresnel integral and I know the answer is 1/2 * (pi / 2)^(1/2), but I'm really getting stuck so I'd be greatful for any help. Thanks In Advance, Kyle Czarnecki ************************ David C. Ullrich David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250 Posted: Sat Jul 08, 2006 8:00 pm    Post subject: Re: Real Integral w/ Complex Analysis On Sat, 08 Jul 2006 11:26:27 EDT, Narcoleptic Insomniac
<i_have_narcoleptic_insomnia@yahoo.com> wrote:

 Quote: On Jul 8, 2006 8:16 AM CT, Narcoleptic Insomniac wrote: All the real integrals I've done with complex analysis have made use of residues (which is one of the coolest things I've seen). Anyways, I've been stumped on the following exercise for awhile which doesn't seem to involve residues. The exercise is to evaluate the integral int_{0}^{oo} cos(x^2) dx ...using complex analysis. I'm supposed consider the real part of e^{iz^2} along a particular contour G. Here a picture would be worth 1000 words, but I'll give my parametrization of G: G = g_1 + g_2 - g_3 g_1: [0, 1] -> C g_1(t) = A * t ...where A is a positive real number, g_2: [0, 1] -> C g_2(t) = A * e^{t * i * pi / 4}, g_3: [0, 1] -> C g_3(t) = A * t * e^{pi * i / 4}. A friend suggested to me a better, more natural parametrization g_1: [0, A] -> C g_1(t) = A

Huh?

 Quote: g_3: [0, A] -> C g_3(t) = t * e^{pi * i / 4} Essentially, G is a pie shaped wedge that begins at the origin, travels along the real axis to the positive number A, goes up in a circular arc pi/4 radians, and then returns to the origin. I'm thinking that since e^{iz^2} is entire we have int_G e^{iz^2} dz = 0 ...by Cauchy's theorem so... int_{g_1} e^{iz^2} dz = int_{g_3 - g_2} e^{iz^2} dz. So if I could evaluate the latter, taking the real part and the limit as A -> oo, then I'd get the value for the cos(x^2) integral. However, when I try do this directly with my parametrization I end up with some nasty integrals. With the better parametrization for g_3 I have int_{g_3} e^{i z^2} dz = e^{pi * i / 4} int_{0, A} e^{-t^2} dt ..so taking the real part of this as A -> oo I get... 1 / sqrt(2) * int_{0, oo} e^{-t^2} dt. Using polar coordinates I can show this last integral to be sqrt(pi) / 2 (is there a better way to do this?) so lim_{A -> oo} Re[int_{g_3} e^{i z^2} dz] = 1/2 * (pi/2)^(1/2). Since this should be the answer all I need to do now is show that lim_{A -> oo} Re[int_{-g_2} e^{i z^2} dz] = 0. I was told that the Jordan lemma would help but I don't see how to apply it in this case.

That "Jordan lemma" thing is incredibly stupid.
Why? Because it's just a special case of a much more
powerful result, _and_ the more general result has
exactly the same simple proof!

Better Lemma: Suppose that f_n:[a,b] -> C is
continuous for each n. Suppose that the f_n
are uniformly bounded, |f_n(x)| <= M for all
n and all x. Suppose that for every d > 0
we have f_n -> 0 uniformly on [a+d, b-d]
as n -> infinity. Then int_a^b f_n -> 0.

This in turn is of course just a special case
of the Dominated Convergence Theorem from
real analysis. But DCT is not quite trivial,
while the Better Lemma _is_ trivial:

Pf: Let eps > 0. Choose d > 0 such that

2dM < eps/2

Choose N so that

|f_n(x)| < eps/(2(b-a))

for all n > N and all x in [a+d, b-d].
Then it follows that

int_a^b |f_n(x)| < eps

for all n > N (split the integral into
two pieces in the obvious way and apply
one inequality above to each piece.) QED.

 Quote: I've read that this is a kind of Fresnel integral and I know the answer is 1/2 * (pi / 2)^(1/2), but I'm really getting stuck so I'd be greatful for any help. Thanks In Advance, Kyle Czarnecki

************************

David C. Ullrich science forum beginner

Joined: 07 May 2006
Posts: 38 Posted: Sat Jul 08, 2006 7:29 pm    Post subject: Re: Real Integral w/ Complex Analysis On Sat, 08 Jul 2006 11:26:27 EDT, Narcoleptic Insomniac
<i_have_narcoleptic_insomnia@yahoo.com> wrote:

[snip]

 Quote: 1 / sqrt(2) * int_{0, oo} e^{-t^2} dt. Using polar coordinates I can show this last integral to be sqrt(pi) / 2 (is there a better way to do this?)

[snip]

FWIW, this page concerning the history of statistics

<http://www.york.ac.uk/depts/maths/histstat/>

references an article titled "Information on the History of the
Normal Law" that gives seven ways of evaluating the probability
integral (trivially different from yours above). Method 7 uses
contour integration. The URL of the referenced PDF file is

<http://www.york.ac.uk/depts/maths/histstat/normal_history.pdf>

There are any number of additional ways to evaluate the Gaussian
integral.

HTH Narcoleptic Insomniac
science forum Guru

Joined: 02 May 2005
Posts: 323 Posted: Sat Jul 08, 2006 3:50 pm    Post subject: Re: Real Integral w/ Complex Analysis On Jul 8, 2006 10:25 AM CT, David C. Ullrich wrote:

 Quote: On Sat, 08 Jul 2006 09:16:49 EDT, Narcoleptic Insomniac wrote: All the real integrals I've done with complex analysis have made use of residues (which is one of the coolest things I've seen). Anyways, I've been stumped on the following exercise for awhile which doesn't seem to involve residues. The exercise is to evaluate the integral int_{0}^{oo} cos(x^2) dx ..using complex analysis. I'm supposed consider the real part of e^{iz^2} along a particular contour G. Here a picture would be worth 1000 words, but I'll give my parametrization of G: G = g_1 + g_2 - g_3 g_1: [0, 1] -> C g_1(t) = A * t ...where A is a positive real number, g_2: [0, 1] -> C g_2(t) = A * e^{t * i * pi / 4}, g_3: [0, 1] -> C g_3(t) = A * t * e^{pi * i / 4}. Essentially, G is a pie shaped wedge that begins at the origin, travels along the real axis to the positive number A, goes up in a circular arc pi/4 radians, and then returns to the origin. I'm thinking that since e^{iz^2} is entire we have int_G e^{iz^2} dz = 0 ..by Cauchy's theorem so... int_{g_1} e^{iz^2} dz = int_{g_3 - g_2} e^{iz^2} dz. So if I could evaluate the latter, taking the real part and the limit as A -> oo, then I'd get the value for the cos(x^2) integral. However, when I try do this directly with my parametrization I end up with some nasty integrals. Well, if you look at the absolute value of the integrand when you integrate over g_2 you should be able to show that it tends to 0 as A -> infinity. Exactly what nasty integral do you get for g_3? Seems to me that it's a nasty but very well-known integral (times a certain constant due to the "dz".)

Yeah, I'm not sure why I couldn't see it right away;
oh well. After I reparametrized g_3 it became more clear
to me (see my other post).

Thanks David.

 Quote: I've read that this is a kind of Fresnel integral and I know the answer is 1/2 * (pi / 2)^(1/2), but I'm really getting stuck so I'd be greatful for any help. Thanks In Advance, Kyle Czarnecki ************************ David C. Ullrich Narcoleptic Insomniac
science forum Guru

Joined: 02 May 2005
Posts: 323 Posted: Sat Jul 08, 2006 3:26 pm    Post subject: Re: Real Integral w/ Complex Analysis On Jul 8, 2006 8:16 AM CT, Narcoleptic Insomniac wrote:

 Quote: All the real integrals I've done with complex analysis have made use of residues (which is one of the coolest things I've seen). Anyways, I've been stumped on the following exercise for awhile which doesn't seem to involve residues. The exercise is to evaluate the integral int_{0}^{oo} cos(x^2) dx ...using complex analysis. I'm supposed consider the real part of e^{iz^2} along a particular contour G. Here a picture would be worth 1000 words, but I'll give my parametrization of G: G = g_1 + g_2 - g_3 g_1: [0, 1] -> C g_1(t) = A * t ...where A is a positive real number, g_2: [0, 1] -> C g_2(t) = A * e^{t * i * pi / 4}, g_3: [0, 1] -> C g_3(t) = A * t * e^{pi * i / 4}.

A friend suggested to me a better, more natural
parametrization

g_1: [0, A] -> C
g_1(t) = A

g_3: [0, A] -> C
g_3(t) = t * e^{pi * i / 4}

 Quote: Essentially, G is a pie shaped wedge that begins at the origin, travels along the real axis to the positive number A, goes up in a circular arc pi/4 radians, and then returns to the origin. I'm thinking that since e^{iz^2} is entire we have int_G e^{iz^2} dz = 0 ...by Cauchy's theorem so... int_{g_1} e^{iz^2} dz = int_{g_3 - g_2} e^{iz^2} dz. So if I could evaluate the latter, taking the real part and the limit as A -> oo, then I'd get the value for the cos(x^2) integral. However, when I try do this directly with my parametrization I end up with some nasty integrals.

With the better parametrization for g_3 I have

int_{g_3} e^{i z^2} dz =

e^{pi * i / 4} int_{0, A} e^{-t^2} dt

...so taking the real part of this as A -> oo I get...

1 / sqrt(2) * int_{0, oo} e^{-t^2} dt.

Using polar coordinates I can show this last integral to
be sqrt(pi) / 2 (is there a better way to do this?) so

lim_{A -> oo} Re[int_{g_3} e^{i z^2} dz] = 1/2 * (pi/2)^(1/2).

Since this should be the answer all I need to do now is
show that

lim_{A -> oo} Re[int_{-g_2} e^{i z^2} dz] = 0.

I was told that the Jordan lemma would help but I don't
see how to apply it in this case.

 Quote: I've read that this is a kind of Fresnel integral and I know the answer is 1/2 * (pi / 2)^(1/2), but I'm really getting stuck so I'd be greatful for any help. Thanks In Advance, Kyle Czarnecki David C. Ullrich
science forum Guru

Joined: 28 Apr 2005
Posts: 2250 Posted: Sat Jul 08, 2006 3:25 pm    Post subject: Re: Real Integral w/ Complex Analysis On Sat, 08 Jul 2006 09:16:49 EDT, Narcoleptic Insomniac
<i_have_narcoleptic_insomnia@yahoo.com> wrote:

 Quote: All the real integrals I've done with complex analysis have made use of residues (which is one of the coolest things I've seen). Anyways, I've been stumped on the following exercise for awhile which doesn't seem to involve residues. The exercise is to evaluate the integral int_{0}^{oo} cos(x^2) dx ..using complex analysis. I'm supposed consider the real part of e^{iz^2} along a particular contour G. Here a picture would be worth 1000 words, but I'll give my parametrization of G: G = g_1 + g_2 - g_3 g_1: [0, 1] -> C g_1(t) = A * t ...where A is a positive real number, g_2: [0, 1] -> C g_2(t) = A * e^{t * i * pi / 4}, g_3: [0, 1] -> C g_3(t) = A * t * e^{pi * i / 4}. Essentially, G is a pie shaped wedge that begins at the origin, travels along the real axis to the positive number A, goes up in a circular arc pi/4 radians, and then returns to the origin. I'm thinking that since e^{iz^2} is entire we have int_G e^{iz^2} dz = 0 ..by Cauchy's theorem so... int_{g_1} e^{iz^2} dz = int_{g_3 - g_2} e^{iz^2} dz. So if I could evaluate the latter, taking the real part and the limit as A -> oo, then I'd get the value for the cos(x^2) integral. However, when I try do this directly with my parametrization I end up with some nasty integrals.

Well, if you look at the absolute value of the integrand
when you integrate over g_2 you should be able to show
that it tends to 0 as A -> infinity.

Exactly what nasty integral do you get for g_3? Seems
to me that it's a nasty but very well-known integral
(times a certain constant due to the "dz".)

 Quote: I've read that this is a kind of Fresnel integral and I know the answer is 1/2 * (pi / 2)^(1/2), but I'm really getting stuck so I'd be greatful for any help. Thanks In Advance, Kyle Czarnecki

************************

David C. Ullrich Narcoleptic Insomniac
science forum Guru

Joined: 02 May 2005
Posts: 323 Posted: Sat Jul 08, 2006 1:16 pm    Post subject: Real Integral w/ Complex Analysis All the real integrals I've done with complex analysis have made use of residues (which is one of the coolest things I've seen). Anyways, I've been stumped on the following exercise for awhile which doesn't seem to involve residues. The exercise is to evaluate the integral int_{0}^{oo} cos(x^2) dx ...using complex analysis. I'm supposed consider the real part of e^{iz^2} along a particular contour G. Here a picture would be worth 1000 words, but I'll give my parametrization of G: G = g_1 + g_2 - g_3 g_1: [0, 1] -> C g_1(t) = A * t ...where A is a positive real number, g_2: [0, 1] -> C g_2(t) = A * e^{t * i * pi / 4}, g_3: [0, 1] -> C g_3(t) = A * t * e^{pi * i / 4}. Essentially, G is a pie shaped wedge that begins at the origin, travels along the real axis to the positive number A, goes up in a circular arc pi/4 radians, and then returns to the origin. I'm thinking that since e^{iz^2} is entire we have int_G e^{iz^2} dz = 0 ...by Cauchy's theorem so... int_{g_1} e^{iz^2} dz = int_{g_3 - g_2} e^{iz^2} dz. So if I could evaluate the latter, taking the real part and the limit as A -> oo, then I'd get the value for the cos(x^2) integral. However, when I try do this directly with my parametrization I end up with some nasty integrals. I've read that this is a kind of Fresnel integral and I know the answer is 1/2 * (pi / 2)^(1/2), but I'm really getting stuck so I'd be greatful for any help. Thanks In Advance, Kyle Czarnecki  Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First
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