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Real Integral w/ Complex Analysis
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David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Sun Jul 09, 2006 12:45 pm    Post subject: Re: Real Integral w/ Complex Analysis Reply with quote

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 <i_have_narcoleptic_insomnia@yahoo.com> 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
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G. A. Edgar
science forum Guru


Joined: 29 Apr 2005
Posts: 470

PostPosted: Sun Jul 09, 2006 12:28 pm    Post subject: Re: Real Integral w/ Complex Analysis Reply with quote

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/
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Narcoleptic Insomniac
science forum Guru


Joined: 02 May 2005
Posts: 323

PostPosted: Sun Jul 09, 2006 5:08 am    Post subject: Re: Real Integral w/ Complex Analysis Reply with quote

On Jul 8, 2006 3:00 PM, David C. Ullrich wrote:

Quote:
On Sat, 08 Jul 2006 11:26:27 EDT, Narcoleptic
Insomniac <i_have_narcoleptic_insomnia@yahoo.com> 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
Back to top
David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Sat Jul 08, 2006 8:00 pm    Post subject: Re: Real Integral w/ Complex Analysis Reply with quote

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
Back to top
Badger
science forum beginner


Joined: 07 May 2006
Posts: 38

PostPosted: Sat Jul 08, 2006 7:29 pm    Post subject: Re: Real Integral w/ Complex Analysis Reply with quote

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
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Narcoleptic Insomniac
science forum Guru


Joined: 02 May 2005
Posts: 323

PostPosted: Sat Jul 08, 2006 3:50 pm    Post subject: Re: Real Integral w/ Complex Analysis Reply with quote

On Jul 8, 2006 10:25 AM CT, David C. Ullrich wrote:

Quote:
On Sat, 08 Jul 2006 09:16:49 EDT, Narcoleptic
Insomniac <i_have_narcoleptic_insomnia@yahoo.com> 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
Back to top
Narcoleptic Insomniac
science forum Guru


Joined: 02 May 2005
Posts: 323

PostPosted: Sat Jul 08, 2006 3:26 pm    Post subject: Re: Real Integral w/ Complex Analysis Reply with quote

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
Back to top
David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Sat Jul 08, 2006 3:25 pm    Post subject: Re: Real Integral w/ Complex Analysis Reply with quote

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
Back to top
Narcoleptic Insomniac
science forum Guru


Joined: 02 May 2005
Posts: 323

PostPosted: Sat Jul 08, 2006 1:16 pm    Post subject: Real Integral w/ Complex Analysis Reply with 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.

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
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