Mixing
An integral I just cannot do
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$$intsum_n=-infty^infty e^-pin^2/ x dx $$
I tried doing this
$$sum_n=-infty^infty int e^-pin^2/ x dx $$ and then
$$sum_n=-infty^infty -pin^2 ln(x) e^(-pin^2/ x) +1over (-pin^2/ x) +1 dx $$ and now I'm not sure what to do..how do I sum this up and if I take the definite integral do I sum it up first or evaluate at $0$ and $infty$
$$int_0^infty sum_n=-infty^infty e^-pin^2/ x dx $$ Does this integral diverge? If so,
What about this one
$$int_0^infty frac 1 sum_n=-infty^infty e^-pin^2/ x dx $$
integration definite-integrals summation
asked Jul 28 at 22:19
Jibran Iqbal
176
add a comment |Â
up vote
-3
down vote
favorite
$$intsum_n=-infty^infty e^-pin^2/ x dx $$
I tried doing this
$$sum_n=-infty^infty int e^-pin^2/ x dx $$ and then
$$sum_n=-infty^infty -pin^2 ln(x) e^(-pin^2/ x) +1over (-pin^2/ x) +1 dx $$ and now I'm not sure what to do..how do I sum this up and if I take the definite integral do I sum it up first or evaluate at $0$ and $infty$
$$int_0^infty sum_n=-infty^infty e^-pin^2/ x dx $$ Does this integral diverge? If so,
What about this one
$$int_0^infty frac 1 sum_n=-infty^infty e^-pin^2/ x dx $$
integration definite-integrals summation
asked Jul 28 at 22:19
Jibran Iqbal
176
No closed form antiderivative for $e^-a/x$. The definite integral diverges.
– Yves Daoust
Jul 28 at 22:25
what about the second one?
– Jibran Iqbal
Jul 29 at 8:49
add a comment |Â
up vote
-3
down vote
favorite
up vote
-3
down vote
favorite
$$intsum_n=-infty^infty e^-pin^2/ x dx $$
I tried doing this
$$sum_n=-infty^infty int e^-pin^2/ x dx $$ and then
$$sum_n=-infty^infty -pin^2 ln(x) e^(-pin^2/ x) +1over (-pin^2/ x) +1 dx $$ and now I'm not sure what to do..how do I sum this up and if I take the definite integral do I sum it up first or evaluate at $0$ and $infty$
$$int_0^infty sum_n=-infty^infty e^-pin^2/ x dx $$ Does this integral diverge? If so,
What about this one
$$int_0^infty frac 1 sum_n=-infty^infty e^-pin^2/ x dx $$
integration definite-integrals summation
asked Jul 28 at 22:19
Jibran Iqbal
176
$$intsum_n=-infty^infty e^-pin^2/ x dx $$
I tried doing this
$$sum_n=-infty^infty int e^-pin^2/ x dx $$ and then
$$sum_n=-infty^infty -pin^2 ln(x) e^(-pin^2/ x) +1over (-pin^2/ x) +1 dx $$ and now I'm not sure what to do..how do I sum this up and if I take the definite integral do I sum it up first or evaluate at $0$ and $infty$
$$int_0^infty sum_n=-infty^infty e^-pin^2/ x dx $$ Does this integral diverge? If so,
What about this one
$$int_0^infty frac 1 sum_n=-infty^infty e^-pin^2/ x dx $$
integration definite-integrals summation
asked Jul 28 at 22:19
Jibran Iqbal
176
edited Jul 29 at 9:12
asked Jul 28 at 22:19
Jibran Iqbal
176
asked Jul 28 at 22:19
Jibran Iqbal
176
asked Jul 28 at 22:19
Jibran Iqbal
176
176
No closed form antiderivative for $e^-a/x$. The definite integral diverges.
– Yves Daoust
Jul 28 at 22:25
what about the second one?
– Jibran Iqbal
Jul 29 at 8:49
add a comment |Â
No closed form antiderivative for $e^-a/x$. The definite integral diverges.
– Yves Daoust
Jul 28 at 22:25
what about the second one?
– Jibran Iqbal
Jul 29 at 8:49
No closed form antiderivative for $e^-a/x$. The definite integral diverges.
– Yves Daoust
Jul 28 at 22:25
No closed form antiderivative for $e^-a/x$. The definite integral diverges.
– Yves Daoust
Jul 28 at 22:25
what about the second one?
– Jibran Iqbal
Jul 29 at 8:49
what about the second one?
– Jibran Iqbal
Jul 29 at 8:49
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
The inner sum can be represented by a theta function:
$$sum_n=-infty^infty q^n^2 = mathcaltheta_3(0,q).$$
See http://mathworld.wolfram.com/JacobiThetaFunctions.html and https://en.wikipedia.org/wiki/Theta_function for documentation.
So
$$sum_n=-infty^infty e^-pi n^2/x = mathcaltheta_3(0,e^-pi/x).$$
The first definite integral you write definitely diverges. This can be seen from the arguments of the other commenters (it's worth noting that they are switching the order of the sum and integral, which is allowed by Tonelli's theorem because all terms of the series are positive). It can also be seen by a quick look at the shape of the function on Wolfram Alpha. I'm not, however, so sure about the second integral.
answered Jul 28 at 23:18
Taylor M
1564
add a comment |Â
up vote
0
down vote
Note that when $x$ is small and negative the argument of the exponential is huge and positive. We can pick one of the terms in the sum, say $n=1$ and feed it to Alpha to be told it does not converge. The integral from $0$ to $infty$ also does not converge because the integrand goes to $1$.
answered Jul 28 at 22:28
Ross Millikan
275k21185351
The definite integral is in the positives.
– Yves Daoust
Jul 28 at 22:29
add a comment |Â
up vote
0
down vote
$$x>epsilonimplies e^-n^2pi/x>e^-n^2pi/epsilon$$
This is sufficient for every term of the summation to cause divergence.
answered Jul 28 at 22:31
Yves Daoust
110k665203
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The inner sum can be represented by a theta function:
$$sum_n=-infty^infty q^n^2 = mathcaltheta_3(0,q).$$
See http://mathworld.wolfram.com/JacobiThetaFunctions.html and https://en.wikipedia.org/wiki/Theta_function for documentation.
So
$$sum_n=-infty^infty e^-pi n^2/x = mathcaltheta_3(0,e^-pi/x).$$
The first definite integral you write definitely diverges. This can be seen from the arguments of the other commenters (it's worth noting that they are switching the order of the sum and integral, which is allowed by Tonelli's theorem because all terms of the series are positive). It can also be seen by a quick look at the shape of the function on Wolfram Alpha. I'm not, however, so sure about the second integral.
answered Jul 28 at 23:18
Taylor M
1564
add a comment |Â
up vote
1
down vote
The inner sum can be represented by a theta function:
$$sum_n=-infty^infty q^n^2 = mathcaltheta_3(0,q).$$
See http://mathworld.wolfram.com/JacobiThetaFunctions.html and https://en.wikipedia.org/wiki/Theta_function for documentation.
So
$$sum_n=-infty^infty e^-pi n^2/x = mathcaltheta_3(0,e^-pi/x).$$
The first definite integral you write definitely diverges. This can be seen from the arguments of the other commenters (it's worth noting that they are switching the order of the sum and integral, which is allowed by Tonelli's theorem because all terms of the series are positive). It can also be seen by a quick look at the shape of the function on Wolfram Alpha. I'm not, however, so sure about the second integral.
answered Jul 28 at 23:18
Taylor M
1564
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The inner sum can be represented by a theta function:
$$sum_n=-infty^infty q^n^2 = mathcaltheta_3(0,q).$$
See http://mathworld.wolfram.com/JacobiThetaFunctions.html and https://en.wikipedia.org/wiki/Theta_function for documentation.
So
$$sum_n=-infty^infty e^-pi n^2/x = mathcaltheta_3(0,e^-pi/x).$$
The first definite integral you write definitely diverges. This can be seen from the arguments of the other commenters (it's worth noting that they are switching the order of the sum and integral, which is allowed by Tonelli's theorem because all terms of the series are positive). It can also be seen by a quick look at the shape of the function on Wolfram Alpha. I'm not, however, so sure about the second integral.
answered Jul 28 at 23:18
Taylor M
1564
The inner sum can be represented by a theta function:
$$sum_n=-infty^infty q^n^2 = mathcaltheta_3(0,q).$$
See http://mathworld.wolfram.com/JacobiThetaFunctions.html and https://en.wikipedia.org/wiki/Theta_function for documentation.
So
$$sum_n=-infty^infty e^-pi n^2/x = mathcaltheta_3(0,e^-pi/x).$$
The first definite integral you write definitely diverges. This can be seen from the arguments of the other commenters (it's worth noting that they are switching the order of the sum and integral, which is allowed by Tonelli's theorem because all terms of the series are positive). It can also be seen by a quick look at the shape of the function on Wolfram Alpha. I'm not, however, so sure about the second integral.
answered Jul 28 at 23:18
Taylor M
1564
answered Jul 28 at 23:18
Taylor M
1564
answered Jul 28 at 23:18
Taylor M
1564
answered Jul 28 at 23:18
Taylor M
1564
1564
add a comment |Â
add a comment |Â
up vote
0
down vote
Note that when $x$ is small and negative the argument of the exponential is huge and positive. We can pick one of the terms in the sum, say $n=1$ and feed it to Alpha to be told it does not converge. The integral from $0$ to $infty$ also does not converge because the integrand goes to $1$.
answered Jul 28 at 22:28
Ross Millikan
275k21185351
The definite integral is in the positives.
– Yves Daoust
Jul 28 at 22:29
add a comment |Â
up vote
0
down vote
Note that when $x$ is small and negative the argument of the exponential is huge and positive. We can pick one of the terms in the sum, say $n=1$ and feed it to Alpha to be told it does not converge. The integral from $0$ to $infty$ also does not converge because the integrand goes to $1$.
answered Jul 28 at 22:28
Ross Millikan
275k21185351
The definite integral is in the positives.
– Yves Daoust
Jul 28 at 22:29
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that when $x$ is small and negative the argument of the exponential is huge and positive. We can pick one of the terms in the sum, say $n=1$ and feed it to Alpha to be told it does not converge. The integral from $0$ to $infty$ also does not converge because the integrand goes to $1$.
answered Jul 28 at 22:28
Ross Millikan
275k21185351
Note that when $x$ is small and negative the argument of the exponential is huge and positive. We can pick one of the terms in the sum, say $n=1$ and feed it to Alpha to be told it does not converge. The integral from $0$ to $infty$ also does not converge because the integrand goes to $1$.
answered Jul 28 at 22:28
Ross Millikan
275k21185351
edited Jul 28 at 22:30
answered Jul 28 at 22:28
Ross Millikan
275k21185351
answered Jul 28 at 22:28
Ross Millikan
275k21185351
answered Jul 28 at 22:28
Ross Millikan
275k21185351
275k21185351
The definite integral is in the positives.
– Yves Daoust
Jul 28 at 22:29
add a comment |Â
The definite integral is in the positives.
– Yves Daoust
Jul 28 at 22:29
The definite integral is in the positives.
– Yves Daoust
Jul 28 at 22:29
The definite integral is in the positives.
– Yves Daoust
Jul 28 at 22:29
add a comment |Â
up vote
0
down vote
$$x>epsilonimplies e^-n^2pi/x>e^-n^2pi/epsilon$$
This is sufficient for every term of the summation to cause divergence.
answered Jul 28 at 22:31
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
$$x>epsilonimplies e^-n^2pi/x>e^-n^2pi/epsilon$$
This is sufficient for every term of the summation to cause divergence.
answered Jul 28 at 22:31
Yves Daoust
110k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$x>epsilonimplies e^-n^2pi/x>e^-n^2pi/epsilon$$
This is sufficient for every term of the summation to cause divergence.
answered Jul 28 at 22:31
Yves Daoust
110k665203
$$x>epsilonimplies e^-n^2pi/x>e^-n^2pi/epsilon$$
This is sufficient for every term of the summation to cause divergence.
answered Jul 28 at 22:31
Yves Daoust
110k665203
edited Jul 28 at 22:37
answered Jul 28 at 22:31
Yves Daoust
110k665203
answered Jul 28 at 22:31
Yves Daoust
110k665203
answered Jul 28 at 22:31
Yves Daoust
110k665203
110k665203
add a comment |Â
add a comment |Â
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No closed form antiderivative for $e^-a/x$. The definite integral diverges.
– Yves Daoust
Jul 28 at 22:25
what about the second one?
– Jibran Iqbal
Jul 29 at 8:49