Line integral depending on a parameter is entire
$begingroup$
Suppose you have a continuous function:
$$phi:[0,1]rightarrow mathbb{C}$$
define the complex function:
$$f(z)=int_0^1phi(t)e^{itz}dt$$
prove that it is entire and calculate it's Taylor expansion centered at $z=0$. Honestly I don't know where to start, I think I have to apply the theorem of holomorphy of a parametric integral but I don't understand how.
Also, how can I apply those results to the sequence of functions:
$$f_n(z)=int_0^n sqrt{t}e^{-tz}dt$$
complex-analysis line-integrals entire-functions
asked Dec 9 '18 at 18:26
Renato FaraoneRenato Faraone
2,33111627
$endgroup$
add a comment |
$begingroup$
Suppose you have a continuous function:
$$phi:[0,1]rightarrow mathbb{C}$$
define the complex function:
$$f(z)=int_0^1phi(t)e^{itz}dt$$
prove that it is entire and calculate it's Taylor expansion centered at $z=0$. Honestly I don't know where to start, I think I have to apply the theorem of holomorphy of a parametric integral but I don't understand how.
Also, how can I apply those results to the sequence of functions:
$$f_n(z)=int_0^n sqrt{t}e^{-tz}dt$$
complex-analysis line-integrals entire-functions
asked Dec 9 '18 at 18:26
Renato FaraoneRenato Faraone
2,33111627
$endgroup$
add a comment |
$begingroup$
Suppose you have a continuous function:
$$phi:[0,1]rightarrow mathbb{C}$$
define the complex function:
$$f(z)=int_0^1phi(t)e^{itz}dt$$
prove that it is entire and calculate it's Taylor expansion centered at $z=0$. Honestly I don't know where to start, I think I have to apply the theorem of holomorphy of a parametric integral but I don't understand how.
Also, how can I apply those results to the sequence of functions:
$$f_n(z)=int_0^n sqrt{t}e^{-tz}dt$$
complex-analysis line-integrals entire-functions
asked Dec 9 '18 at 18:26
Renato FaraoneRenato Faraone
2,33111627
$endgroup$
Suppose you have a continuous function:
$$phi:[0,1]rightarrow mathbb{C}$$
define the complex function:
$$f(z)=int_0^1phi(t)e^{itz}dt$$
prove that it is entire and calculate it's Taylor expansion centered at $z=0$. Honestly I don't know where to start, I think I have to apply the theorem of holomorphy of a parametric integral but I don't understand how.
Also, how can I apply those results to the sequence of functions:
$$f_n(z)=int_0^n sqrt{t}e^{-tz}dt$$
complex-analysis line-integrals entire-functions
complex-analysis line-integrals entire-functions
asked Dec 9 '18 at 18:26
Renato FaraoneRenato Faraone
2,33111627
asked Dec 9 '18 at 18:26
Renato FaraoneRenato Faraone
2,33111627
edited Dec 10 '18 at 10:30
Renato Faraone
asked Dec 9 '18 at 18:26
Renato FaraoneRenato Faraone
2,33111627
asked Dec 9 '18 at 18:26
Renato FaraoneRenato Faraone
2,33111627
asked Dec 9 '18 at 18:26
Renato FaraoneRenato Faraone
2,33111627
2,33111627
add a comment |
add a comment |
1 Answer
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$begingroup$
It's obviously entire as the derivative is $$ int_0^1 ite^{izt} phi(t) , dt$$ which exists (is convergent) for all $z$ as the integrand is bounded.
The Taylor series is obtained by differentiating under the integral sign: the $n$th derivative is $$ int_0^1 (it)^n phi(t), dt.$$
answered Dec 9 '18 at 18:30
Richard MartinRichard Martin
1,61118
$endgroup$
$begingroup$
What do you mean by "the integral is convergent"?
$endgroup$
– Renato Faraone
Dec 9 '18 at 18:39
$begingroup$
I mean it exists as a Riemann integral.
$endgroup$
– Richard Martin
Dec 9 '18 at 18:42
$begingroup$
You should probably justify differentiating under the integral sign....
$endgroup$
– qbert
Dec 9 '18 at 19:37
1
$begingroup$
@RichardMartin I don't understand your point. In any event, while the exercise might be straightforward or obvious to you, it probably isn't to the OP (indeed, it is an exercise for a reason). So if you are sweeping things under the rug, you should at least say so
$endgroup$
– qbert
Dec 9 '18 at 20:34
1
$begingroup$
@qbert I agree, every exercise is straightforward if you know how to do it and which theorem to apply. Being this my first course centered on complex analysis I obviously have a hard time even on some basic facts.
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:18
|
show 1 more comment
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
It's obviously entire as the derivative is $$ int_0^1 ite^{izt} phi(t) , dt$$ which exists (is convergent) for all $z$ as the integrand is bounded.
The Taylor series is obtained by differentiating under the integral sign: the $n$th derivative is $$ int_0^1 (it)^n phi(t), dt.$$
answered Dec 9 '18 at 18:30
Richard MartinRichard Martin
1,61118
$endgroup$
$begingroup$
What do you mean by "the integral is convergent"?
$endgroup$
– Renato Faraone
Dec 9 '18 at 18:39
$begingroup$
I mean it exists as a Riemann integral.
$endgroup$
– Richard Martin
Dec 9 '18 at 18:42
$begingroup$
You should probably justify differentiating under the integral sign....
$endgroup$
– qbert
Dec 9 '18 at 19:37
1
$begingroup$
@RichardMartin I don't understand your point. In any event, while the exercise might be straightforward or obvious to you, it probably isn't to the OP (indeed, it is an exercise for a reason). So if you are sweeping things under the rug, you should at least say so
$endgroup$
– qbert
Dec 9 '18 at 20:34
1
$begingroup$
@qbert I agree, every exercise is straightforward if you know how to do it and which theorem to apply. Being this my first course centered on complex analysis I obviously have a hard time even on some basic facts.
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:18
|
show 1 more comment
$begingroup$
It's obviously entire as the derivative is $$ int_0^1 ite^{izt} phi(t) , dt$$ which exists (is convergent) for all $z$ as the integrand is bounded.
The Taylor series is obtained by differentiating under the integral sign: the $n$th derivative is $$ int_0^1 (it)^n phi(t), dt.$$
answered Dec 9 '18 at 18:30
Richard MartinRichard Martin
1,61118
$endgroup$
$begingroup$
What do you mean by "the integral is convergent"?
$endgroup$
– Renato Faraone
Dec 9 '18 at 18:39
$begingroup$
I mean it exists as a Riemann integral.
$endgroup$
– Richard Martin
Dec 9 '18 at 18:42
$begingroup$
You should probably justify differentiating under the integral sign....
$endgroup$
– qbert
Dec 9 '18 at 19:37
1
$begingroup$
@RichardMartin I don't understand your point. In any event, while the exercise might be straightforward or obvious to you, it probably isn't to the OP (indeed, it is an exercise for a reason). So if you are sweeping things under the rug, you should at least say so
$endgroup$
– qbert
Dec 9 '18 at 20:34
1
$begingroup$
@qbert I agree, every exercise is straightforward if you know how to do it and which theorem to apply. Being this my first course centered on complex analysis I obviously have a hard time even on some basic facts.
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:18
|
show 1 more comment
$begingroup$
It's obviously entire as the derivative is $$ int_0^1 ite^{izt} phi(t) , dt$$ which exists (is convergent) for all $z$ as the integrand is bounded.
The Taylor series is obtained by differentiating under the integral sign: the $n$th derivative is $$ int_0^1 (it)^n phi(t), dt.$$
answered Dec 9 '18 at 18:30
Richard MartinRichard Martin
1,61118
$endgroup$
It's obviously entire as the derivative is $$ int_0^1 ite^{izt} phi(t) , dt$$ which exists (is convergent) for all $z$ as the integrand is bounded.
The Taylor series is obtained by differentiating under the integral sign: the $n$th derivative is $$ int_0^1 (it)^n phi(t), dt.$$
answered Dec 9 '18 at 18:30
Richard MartinRichard Martin
1,61118
edited Dec 9 '18 at 19:33
answered Dec 9 '18 at 18:30
Richard MartinRichard Martin
1,61118
answered Dec 9 '18 at 18:30
Richard MartinRichard Martin
1,61118
answered Dec 9 '18 at 18:30
Richard MartinRichard Martin
1,61118
1,61118
$begingroup$
What do you mean by "the integral is convergent"?
$endgroup$
– Renato Faraone
Dec 9 '18 at 18:39
$begingroup$
I mean it exists as a Riemann integral.
$endgroup$
– Richard Martin
Dec 9 '18 at 18:42
$begingroup$
You should probably justify differentiating under the integral sign....
$endgroup$
– qbert
Dec 9 '18 at 19:37
1
$begingroup$
@RichardMartin I don't understand your point. In any event, while the exercise might be straightforward or obvious to you, it probably isn't to the OP (indeed, it is an exercise for a reason). So if you are sweeping things under the rug, you should at least say so
$endgroup$
– qbert
Dec 9 '18 at 20:34
1
$begingroup$
@qbert I agree, every exercise is straightforward if you know how to do it and which theorem to apply. Being this my first course centered on complex analysis I obviously have a hard time even on some basic facts.
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:18
|
show 1 more comment
$begingroup$
What do you mean by "the integral is convergent"?
$endgroup$
– Renato Faraone
Dec 9 '18 at 18:39
$begingroup$
I mean it exists as a Riemann integral.
$endgroup$
– Richard Martin
Dec 9 '18 at 18:42
$begingroup$
You should probably justify differentiating under the integral sign....
$endgroup$
– qbert
Dec 9 '18 at 19:37
1
$begingroup$
@RichardMartin I don't understand your point. In any event, while the exercise might be straightforward or obvious to you, it probably isn't to the OP (indeed, it is an exercise for a reason). So if you are sweeping things under the rug, you should at least say so
$endgroup$
– qbert
Dec 9 '18 at 20:34
1
$begingroup$
@qbert I agree, every exercise is straightforward if you know how to do it and which theorem to apply. Being this my first course centered on complex analysis I obviously have a hard time even on some basic facts.
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:18
$begingroup$
What do you mean by "the integral is convergent"?
$endgroup$
– Renato Faraone
Dec 9 '18 at 18:39
$begingroup$
What do you mean by "the integral is convergent"?
$endgroup$
– Renato Faraone
Dec 9 '18 at 18:39
$begingroup$
I mean it exists as a Riemann integral.
$endgroup$
– Richard Martin
Dec 9 '18 at 18:42
$begingroup$
I mean it exists as a Riemann integral.
$endgroup$
– Richard Martin
Dec 9 '18 at 18:42
$begingroup$
You should probably justify differentiating under the integral sign....
$endgroup$
– qbert
Dec 9 '18 at 19:37
$begingroup$
You should probably justify differentiating under the integral sign....
$endgroup$
– qbert
Dec 9 '18 at 19:37
1
1
$begingroup$
@RichardMartin I don't understand your point. In any event, while the exercise might be straightforward or obvious to you, it probably isn't to the OP (indeed, it is an exercise for a reason). So if you are sweeping things under the rug, you should at least say so
$endgroup$
– qbert
Dec 9 '18 at 20:34
$begingroup$
@RichardMartin I don't understand your point. In any event, while the exercise might be straightforward or obvious to you, it probably isn't to the OP (indeed, it is an exercise for a reason). So if you are sweeping things under the rug, you should at least say so
$endgroup$
– qbert
Dec 9 '18 at 20:34
1
1
$begingroup$
@qbert I agree, every exercise is straightforward if you know how to do it and which theorem to apply. Being this my first course centered on complex analysis I obviously have a hard time even on some basic facts.
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:18
$begingroup$
@qbert I agree, every exercise is straightforward if you know how to do it and which theorem to apply. Being this my first course centered on complex analysis I obviously have a hard time even on some basic facts.
$endgroup$
– Renato Faraone
Dec 10 '18 at 8:18
|
show 1 more comment
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