Improper integral with a parameter $int_{0}^{infty} e^{-cx^{2}}sin(tx)dx$
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I need to evaluate the following integral.
$$int_{0}^{infty} e^{-cx^{2}}sin(tx) ~dx$$
Here's what I've done so far.
$$I(t) = int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$
Then differentiating under the integral sign I get:
$$I'(t) = int_{0}^{infty} xe^{-cx^{2}}cos(tx)~dx$$
This is pretty straightforward and integrating by parts I get this:
$$I '(t) =frac{1}{2c}+frac{t}{2c}int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$ which can be written as
$$I'(t) = frac{1}{2c}+frac{t}{2c}I(t)$$
Rearranging I end up with the following differential equation:
$$I'(t)+frac{t}{2c}I(t)=frac{1}{2c}$$
When I try to solve this I end up with a different result than what an online calculator got:
$$-dfrac{sqrt{{pi}}mathrm{i}mathrm{e}^{-frac{t^2}{4c}}operatorname{erf}left(frac{mathrm{i}t}{2sqrt{c}}right)}{2sqrt{c}}$$
Any tips?
calculus integration improper-integrals
edited Dec 2 at 13:37
Tianlalu
2,9801936
asked Dec 1 at 4:23
Louis Lane
113
New contributor
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
2
down vote
favorite
I need to evaluate the following integral.
$$int_{0}^{infty} e^{-cx^{2}}sin(tx) ~dx$$
Here's what I've done so far.
$$I(t) = int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$
Then differentiating under the integral sign I get:
$$I'(t) = int_{0}^{infty} xe^{-cx^{2}}cos(tx)~dx$$
This is pretty straightforward and integrating by parts I get this:
$$I '(t) =frac{1}{2c}+frac{t}{2c}int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$ which can be written as
$$I'(t) = frac{1}{2c}+frac{t}{2c}I(t)$$
Rearranging I end up with the following differential equation:
$$I'(t)+frac{t}{2c}I(t)=frac{1}{2c}$$
When I try to solve this I end up with a different result than what an online calculator got:
$$-dfrac{sqrt{{pi}}mathrm{i}mathrm{e}^{-frac{t^2}{4c}}operatorname{erf}left(frac{mathrm{i}t}{2sqrt{c}}right)}{2sqrt{c}}$$
Any tips?
calculus integration improper-integrals
edited Dec 2 at 13:37
Tianlalu
2,9801936
asked Dec 1 at 4:23
Louis Lane
113
New contributor
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You must've also ended with same answer. This is a linear first oder differential equation you ended up with. Now multiplying both sides by $e^{frac {t^2}{4c}}$,LHS can be solved much easily but on the right side since you need an indefinite integral of type $e^{-x^2}$ you must obviously end with something containing the error function within.
– Digamma
Dec 1 at 12:27
The online calculator answer is correct! It may also be found in published tables of integrals. (You do not reveal to us your answer, so we cannot tell whether it is also correct.)
– GEdgar
Dec 1 at 14:02
Your error lies in differential equation that you've formed. It should be $I'(t) - frac{t}{2c}I(t) = frac{1}{2c}$
– DavidG
2 days ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to evaluate the following integral.
$$int_{0}^{infty} e^{-cx^{2}}sin(tx) ~dx$$
Here's what I've done so far.
$$I(t) = int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$
Then differentiating under the integral sign I get:
$$I'(t) = int_{0}^{infty} xe^{-cx^{2}}cos(tx)~dx$$
This is pretty straightforward and integrating by parts I get this:
$$I '(t) =frac{1}{2c}+frac{t}{2c}int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$ which can be written as
$$I'(t) = frac{1}{2c}+frac{t}{2c}I(t)$$
Rearranging I end up with the following differential equation:
$$I'(t)+frac{t}{2c}I(t)=frac{1}{2c}$$
When I try to solve this I end up with a different result than what an online calculator got:
$$-dfrac{sqrt{{pi}}mathrm{i}mathrm{e}^{-frac{t^2}{4c}}operatorname{erf}left(frac{mathrm{i}t}{2sqrt{c}}right)}{2sqrt{c}}$$
Any tips?
calculus integration improper-integrals
edited Dec 2 at 13:37
Tianlalu
2,9801936
asked Dec 1 at 4:23
Louis Lane
113
New contributor
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I need to evaluate the following integral.
$$int_{0}^{infty} e^{-cx^{2}}sin(tx) ~dx$$
Here's what I've done so far.
$$I(t) = int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$
Then differentiating under the integral sign I get:
$$I'(t) = int_{0}^{infty} xe^{-cx^{2}}cos(tx)~dx$$
This is pretty straightforward and integrating by parts I get this:
$$I '(t) =frac{1}{2c}+frac{t}{2c}int_{0}^{infty} e^{-cx^{2}}sin(tx)~dx$$ which can be written as
$$I'(t) = frac{1}{2c}+frac{t}{2c}I(t)$$
Rearranging I end up with the following differential equation:
$$I'(t)+frac{t}{2c}I(t)=frac{1}{2c}$$
When I try to solve this I end up with a different result than what an online calculator got:
$$-dfrac{sqrt{{pi}}mathrm{i}mathrm{e}^{-frac{t^2}{4c}}operatorname{erf}left(frac{mathrm{i}t}{2sqrt{c}}right)}{2sqrt{c}}$$
Any tips?
calculus integration improper-integrals
calculus integration improper-integrals
edited Dec 2 at 13:37
Tianlalu
2,9801936
asked Dec 1 at 4:23
Louis Lane
113
New contributor
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 13:37
Tianlalu
2,9801936
asked Dec 1 at 4:23
Louis Lane
113
New contributor
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 2 at 13:37
Tianlalu
2,9801936
edited Dec 2 at 13:37
Tianlalu
2,9801936
edited Dec 2 at 13:37
Tianlalu
2,9801936
2,9801936
asked Dec 1 at 4:23
Louis Lane
113
New contributor
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 1 at 4:23
Louis Lane
113
asked Dec 1 at 4:23
Louis Lane
113
113
New contributor
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Louis Lane is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You must've also ended with same answer. This is a linear first oder differential equation you ended up with. Now multiplying both sides by $e^{frac {t^2}{4c}}$,LHS can be solved much easily but on the right side since you need an indefinite integral of type $e^{-x^2}$ you must obviously end with something containing the error function within.
– Digamma
Dec 1 at 12:27
The online calculator answer is correct! It may also be found in published tables of integrals. (You do not reveal to us your answer, so we cannot tell whether it is also correct.)
– GEdgar
Dec 1 at 14:02
Your error lies in differential equation that you've formed. It should be $I'(t) - frac{t}{2c}I(t) = frac{1}{2c}$
– DavidG
2 days ago
add a comment |
You must've also ended with same answer. This is a linear first oder differential equation you ended up with. Now multiplying both sides by $e^{frac {t^2}{4c}}$,LHS can be solved much easily but on the right side since you need an indefinite integral of type $e^{-x^2}$ you must obviously end with something containing the error function within.
– Digamma
Dec 1 at 12:27
The online calculator answer is correct! It may also be found in published tables of integrals. (You do not reveal to us your answer, so we cannot tell whether it is also correct.)
– GEdgar
Dec 1 at 14:02
Your error lies in differential equation that you've formed. It should be $I'(t) - frac{t}{2c}I(t) = frac{1}{2c}$
– DavidG
2 days ago
You must've also ended with same answer. This is a linear first oder differential equation you ended up with. Now multiplying both sides by $e^{frac {t^2}{4c}}$,LHS can be solved much easily but on the right side since you need an indefinite integral of type $e^{-x^2}$ you must obviously end with something containing the error function within.
– Digamma
Dec 1 at 12:27
You must've also ended with same answer. This is a linear first oder differential equation you ended up with. Now multiplying both sides by $e^{frac {t^2}{4c}}$,LHS can be solved much easily but on the right side since you need an indefinite integral of type $e^{-x^2}$ you must obviously end with something containing the error function within.
– Digamma
Dec 1 at 12:27
The online calculator answer is correct! It may also be found in published tables of integrals. (You do not reveal to us your answer, so we cannot tell whether it is also correct.)
– GEdgar
Dec 1 at 14:02
The online calculator answer is correct! It may also be found in published tables of integrals. (You do not reveal to us your answer, so we cannot tell whether it is also correct.)
– GEdgar
Dec 1 at 14:02
Your error lies in differential equation that you've formed. It should be $I'(t) - frac{t}{2c}I(t) = frac{1}{2c}$
– DavidG
2 days ago
Your error lies in differential equation that you've formed. It should be $I'(t) - frac{t}{2c}I(t) = frac{1}{2c}$
– DavidG
2 days ago
add a comment |
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Louis Lane is a new contributor. Be nice, and check out our Code of Conduct.
Louis Lane is a new contributor. Be nice, and check out our Code of Conduct.
Louis Lane is a new contributor. Be nice, and check out our Code of Conduct.
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You must've also ended with same answer. This is a linear first oder differential equation you ended up with. Now multiplying both sides by $e^{frac {t^2}{4c}}$,LHS can be solved much easily but on the right side since you need an indefinite integral of type $e^{-x^2}$ you must obviously end with something containing the error function within.
– Digamma
Dec 1 at 12:27
The online calculator answer is correct! It may also be found in published tables of integrals. (You do not reveal to us your answer, so we cannot tell whether it is also correct.)
– GEdgar
Dec 1 at 14:02
Your error lies in differential equation that you've formed. It should be $I'(t) - frac{t}{2c}I(t) = frac{1}{2c}$
– DavidG
2 days ago