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Problem using Parseval's theorem for solving an integral
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I need to use Parseval's theorem to calculate the following integral:
$$int_-infty^inftyleft |frac1-e^-iwtiw right |^2dt$$
I thought to find the transform of $$f(t) = frac1-e^-iwtiw$$ and then use Parseval.
I have problems finding the Fourier transform of f(t) using properties by knowing common transforms.
The definition used for the Fourier transform is:
$$hatf(w) = int_-infty^infty f(t) e^-iwtcdot dt$$
calculus improper-integrals fourier-transform parsevals-identity
edited Jul 27 at 18:18
Math Lover
12.3k21232
asked Jul 27 at 16:14
Verónica
235
add a comment |Â
up vote
2
down vote
favorite
I need to use Parseval's theorem to calculate the following integral:
$$int_-infty^inftyleft |frac1-e^-iwtiw right |^2dt$$
I thought to find the transform of $$f(t) = frac1-e^-iwtiw$$ and then use Parseval.
I have problems finding the Fourier transform of f(t) using properties by knowing common transforms.
The definition used for the Fourier transform is:
$$hatf(w) = int_-infty^infty f(t) e^-iwtcdot dt$$
calculus improper-integrals fourier-transform parsevals-identity
edited Jul 27 at 18:18
Math Lover
12.3k21232
asked Jul 27 at 16:14
Verónica
235
1
It might be useful to note that $f(t) = frac1-e^-iwtiw = int_0^t e^-iwu , du$, so $f'(t) = e^-iwu$.
– Sobi
Jul 27 at 16:30
1
Be careful, the $w$ in $f(t)$ is not the same $w$ in $hat f(w).$
– Thomas Andrews
Jul 27 at 16:33
Thank you! Shouldn't it be the integral of the exponential multiplied by the Heavyside function from -oo to t? If that's correct then I should find f(w)/(iw) to calculate the integral by Parseval right?
– Verónica
Jul 27 at 16:48
Perhaps I'm missing something, but the integrand is positive and periodic and integrating a positive, periodic function over all $mathbbR$ usually diverges.
– robjohn♦
Jul 27 at 18:25
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to use Parseval's theorem to calculate the following integral:
$$int_-infty^inftyleft |frac1-e^-iwtiw right |^2dt$$
I thought to find the transform of $$f(t) = frac1-e^-iwtiw$$ and then use Parseval.
I have problems finding the Fourier transform of f(t) using properties by knowing common transforms.
The definition used for the Fourier transform is:
$$hatf(w) = int_-infty^infty f(t) e^-iwtcdot dt$$
calculus improper-integrals fourier-transform parsevals-identity
edited Jul 27 at 18:18
Math Lover
12.3k21232
asked Jul 27 at 16:14
Verónica
235
I need to use Parseval's theorem to calculate the following integral:
$$int_-infty^inftyleft |frac1-e^-iwtiw right |^2dt$$
I thought to find the transform of $$f(t) = frac1-e^-iwtiw$$ and then use Parseval.
I have problems finding the Fourier transform of f(t) using properties by knowing common transforms.
The definition used for the Fourier transform is:
$$hatf(w) = int_-infty^infty f(t) e^-iwtcdot dt$$
calculus improper-integrals fourier-transform parsevals-identity
edited Jul 27 at 18:18
Math Lover
12.3k21232
asked Jul 27 at 16:14
Verónica
235
edited Jul 27 at 18:18
Math Lover
12.3k21232
edited Jul 27 at 18:18
Math Lover
12.3k21232
edited Jul 27 at 18:18
Math Lover
12.3k21232
12.3k21232
asked Jul 27 at 16:14
Verónica
235
asked Jul 27 at 16:14
Verónica
235
asked Jul 27 at 16:14
Verónica
235
235
1
It might be useful to note that $f(t) = frac1-e^-iwtiw = int_0^t e^-iwu , du$, so $f'(t) = e^-iwu$.
– Sobi
Jul 27 at 16:30
1
Be careful, the $w$ in $f(t)$ is not the same $w$ in $hat f(w).$
– Thomas Andrews
Jul 27 at 16:33
Thank you! Shouldn't it be the integral of the exponential multiplied by the Heavyside function from -oo to t? If that's correct then I should find f(w)/(iw) to calculate the integral by Parseval right?
– Verónica
Jul 27 at 16:48
Perhaps I'm missing something, but the integrand is positive and periodic and integrating a positive, periodic function over all $mathbbR$ usually diverges.
– robjohn♦
Jul 27 at 18:25
add a comment |Â
1
It might be useful to note that $f(t) = frac1-e^-iwtiw = int_0^t e^-iwu , du$, so $f'(t) = e^-iwu$.
– Sobi
Jul 27 at 16:30
1
Be careful, the $w$ in $f(t)$ is not the same $w$ in $hat f(w).$
– Thomas Andrews
Jul 27 at 16:33
Thank you! Shouldn't it be the integral of the exponential multiplied by the Heavyside function from -oo to t? If that's correct then I should find f(w)/(iw) to calculate the integral by Parseval right?
– Verónica
Jul 27 at 16:48
Perhaps I'm missing something, but the integrand is positive and periodic and integrating a positive, periodic function over all $mathbbR$ usually diverges.
– robjohn♦
Jul 27 at 18:25
1
1
It might be useful to note that $f(t) = frac1-e^-iwtiw = int_0^t e^-iwu , du$, so $f'(t) = e^-iwu$.
– Sobi
Jul 27 at 16:30
It might be useful to note that $f(t) = frac1-e^-iwtiw = int_0^t e^-iwu , du$, so $f'(t) = e^-iwu$.
– Sobi
Jul 27 at 16:30
1
1
Be careful, the $w$ in $f(t)$ is not the same $w$ in $hat f(w).$
– Thomas Andrews
Jul 27 at 16:33
Be careful, the $w$ in $f(t)$ is not the same $w$ in $hat f(w).$
– Thomas Andrews
Jul 27 at 16:33
Thank you! Shouldn't it be the integral of the exponential multiplied by the Heavyside function from -oo to t? If that's correct then I should find f(w)/(iw) to calculate the integral by Parseval right?
– Verónica
Jul 27 at 16:48
Thank you! Shouldn't it be the integral of the exponential multiplied by the Heavyside function from -oo to t? If that's correct then I should find f(w)/(iw) to calculate the integral by Parseval right?
– Verónica
Jul 27 at 16:48
Perhaps I'm missing something, but the integrand is positive and periodic and integrating a positive, periodic function over all $mathbbR$ usually diverges.
– robjohn♦
Jul 27 at 18:25
Perhaps I'm missing something, but the integrand is positive and periodic and integrating a positive, periodic function over all $mathbbR$ usually diverges.
– robjohn♦
Jul 27 at 18:25
add a comment |Â
1 Answer
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1
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Hint: I think the denominator should be $wt$ instead of $w$ as the integral would evaluate to $infty$ otherwise. Also,
$$frac1-e^-iwtiwt = e^-iwt/2frace^iwt/2-e^-iwt/2iwt = e^-iwt/2fracsin(wt/2)wt/2.$$
Consequently,
$$left|frac1-e^-iwtiwtright| = left|fracsin(wt/2)wt/2right|.$$
answered Jul 27 at 17:16
Math Lover
12.3k21232
Yes that should be great but it's not iwt, its iw.
– Verónica
Jul 27 at 17:18
@Verónica In that case, as I also mentioned in the answer, the integral would evaluate to $infty$ as $sin^2(theta) = colorredfrac12-0.5cos(2theta)$.
– Math Lover
Jul 27 at 17:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint: I think the denominator should be $wt$ instead of $w$ as the integral would evaluate to $infty$ otherwise. Also,
$$frac1-e^-iwtiwt = e^-iwt/2frace^iwt/2-e^-iwt/2iwt = e^-iwt/2fracsin(wt/2)wt/2.$$
Consequently,
$$left|frac1-e^-iwtiwtright| = left|fracsin(wt/2)wt/2right|.$$
answered Jul 27 at 17:16
Math Lover
12.3k21232
Yes that should be great but it's not iwt, its iw.
– Verónica
Jul 27 at 17:18
@Verónica In that case, as I also mentioned in the answer, the integral would evaluate to $infty$ as $sin^2(theta) = colorredfrac12-0.5cos(2theta)$.
– Math Lover
Jul 27 at 17:21
add a comment |Â
up vote
1
down vote
Hint: I think the denominator should be $wt$ instead of $w$ as the integral would evaluate to $infty$ otherwise. Also,
$$frac1-e^-iwtiwt = e^-iwt/2frace^iwt/2-e^-iwt/2iwt = e^-iwt/2fracsin(wt/2)wt/2.$$
Consequently,
$$left|frac1-e^-iwtiwtright| = left|fracsin(wt/2)wt/2right|.$$
answered Jul 27 at 17:16
Math Lover
12.3k21232
Yes that should be great but it's not iwt, its iw.
– Verónica
Jul 27 at 17:18
@Verónica In that case, as I also mentioned in the answer, the integral would evaluate to $infty$ as $sin^2(theta) = colorredfrac12-0.5cos(2theta)$.
– Math Lover
Jul 27 at 17:21
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: I think the denominator should be $wt$ instead of $w$ as the integral would evaluate to $infty$ otherwise. Also,
$$frac1-e^-iwtiwt = e^-iwt/2frace^iwt/2-e^-iwt/2iwt = e^-iwt/2fracsin(wt/2)wt/2.$$
Consequently,
$$left|frac1-e^-iwtiwtright| = left|fracsin(wt/2)wt/2right|.$$
answered Jul 27 at 17:16
Math Lover
12.3k21232
Hint: I think the denominator should be $wt$ instead of $w$ as the integral would evaluate to $infty$ otherwise. Also,
$$frac1-e^-iwtiwt = e^-iwt/2frace^iwt/2-e^-iwt/2iwt = e^-iwt/2fracsin(wt/2)wt/2.$$
Consequently,
$$left|frac1-e^-iwtiwtright| = left|fracsin(wt/2)wt/2right|.$$
answered Jul 27 at 17:16
Math Lover
12.3k21232
answered Jul 27 at 17:16
Math Lover
12.3k21232
answered Jul 27 at 17:16
Math Lover
12.3k21232
answered Jul 27 at 17:16
Math Lover
12.3k21232
12.3k21232
Yes that should be great but it's not iwt, its iw.
– Verónica
Jul 27 at 17:18
@Verónica In that case, as I also mentioned in the answer, the integral would evaluate to $infty$ as $sin^2(theta) = colorredfrac12-0.5cos(2theta)$.
– Math Lover
Jul 27 at 17:21
add a comment |Â
Yes that should be great but it's not iwt, its iw.
– Verónica
Jul 27 at 17:18
@Verónica In that case, as I also mentioned in the answer, the integral would evaluate to $infty$ as $sin^2(theta) = colorredfrac12-0.5cos(2theta)$.
– Math Lover
Jul 27 at 17:21
Yes that should be great but it's not iwt, its iw.
– Verónica
Jul 27 at 17:18
Yes that should be great but it's not iwt, its iw.
– Verónica
Jul 27 at 17:18
@Verónica In that case, as I also mentioned in the answer, the integral would evaluate to $infty$ as $sin^2(theta) = colorredfrac12-0.5cos(2theta)$.
– Math Lover
Jul 27 at 17:21
@Verónica In that case, as I also mentioned in the answer, the integral would evaluate to $infty$ as $sin^2(theta) = colorredfrac12-0.5cos(2theta)$.
– Math Lover
Jul 27 at 17:21
add a comment |Â
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1
It might be useful to note that $f(t) = frac1-e^-iwtiw = int_0^t e^-iwu , du$, so $f'(t) = e^-iwu$.
– Sobi
Jul 27 at 16:30
1
Be careful, the $w$ in $f(t)$ is not the same $w$ in $hat f(w).$
– Thomas Andrews
Jul 27 at 16:33
Thank you! Shouldn't it be the integral of the exponential multiplied by the Heavyside function from -oo to t? If that's correct then I should find f(w)/(iw) to calculate the integral by Parseval right?
– Verónica
Jul 27 at 16:48
Perhaps I'm missing something, but the integrand is positive and periodic and integrating a positive, periodic function over all $mathbbR$ usually diverges.
– robjohn♦
Jul 27 at 18:25