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Integration for Advanced Learners [duplicate]
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Proving $int_0^infty mathrme^-x^2 dx = fracsqrt pi2$
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I am a Young leaner of Mathematics in the university and I found this problem a bit challenging et me.
Please can someone help me integrate
$$int_-infty^infty e^-x^-2 dx$$
Thanks.
definite-integrals improper-integrals complex-integration
asked Jul 29 at 13:17
Jordan Fondja
11
marked as duplicate by Hans Lundmark, user 108128, Batominovski, Mark Viola, Jack D'Aurizio♦ definite-integrals
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This question already has an answer here:
Proving $int_0^infty mathrme^-x^2 dx = fracsqrt pi2$
19 answers
I am a Young leaner of Mathematics in the university and I found this problem a bit challenging et me.
Please can someone help me integrate
$$int_-infty^infty e^-x^-2 dx$$
Thanks.
definite-integrals improper-integrals complex-integration
asked Jul 29 at 13:17
Jordan Fondja
11
marked as duplicate by Hans Lundmark, user 108128, Batominovski, Mark Viola, Jack D'Aurizio♦ definite-integrals
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Jul 29 at 21:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
@Sobi this is actually quite challenging for a beginner...
– Karn Watcharasupat
Jul 29 at 13:20
I tried to use integration by part. But it never finishes. Then when I applied the limits, it turns to pi
– Jordan Fondja
Jul 29 at 13:20
@JordanFondja this integral doesn't converge actually. If you plot the integrand out, you will notice that $e^-x^-2to1$ as $xtopminfty$ so the area under the graph till infinity can't be found.
– Karn Watcharasupat
Jul 29 at 13:23
@KarnWatcharasupat I am well aware of that. Just wanted to see how he thinks.
– Sobi
Jul 29 at 13:23
2
@KarnWatcharasupat No. It is x To the power -2
– Jordan Fondja
Jul 29 at 13:42
 |Â
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up vote
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down vote
favorite
This question already has an answer here:
Proving $int_0^infty mathrme^-x^2 dx = fracsqrt pi2$
19 answers
I am a Young leaner of Mathematics in the university and I found this problem a bit challenging et me.
Please can someone help me integrate
$$int_-infty^infty e^-x^-2 dx$$
Thanks.
definite-integrals improper-integrals complex-integration
asked Jul 29 at 13:17
Jordan Fondja
11
This question already has an answer here:
Proving $int_0^infty mathrme^-x^2 dx = fracsqrt pi2$
19 answers
I am a Young leaner of Mathematics in the university and I found this problem a bit challenging et me.
Please can someone help me integrate
$$int_-infty^infty e^-x^-2 dx$$
Thanks.
This question already has an answer here:
Proving $int_0^infty mathrme^-x^2 dx = fracsqrt pi2$
19 answers
definite-integrals improper-integrals complex-integration
asked Jul 29 at 13:17
Jordan Fondja
11
edited Jul 31 at 9:48
asked Jul 29 at 13:17
Jordan Fondja
11
asked Jul 29 at 13:17
Jordan Fondja
11
asked Jul 29 at 13:17
Jordan Fondja
11
11
marked as duplicate by Hans Lundmark, user 108128, Batominovski, Mark Viola, Jack D'Aurizio♦ definite-integrals
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marked as duplicate by Hans Lundmark, user 108128, Batominovski, Mark Viola, Jack D'Aurizio♦ definite-integrals
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Jul 29 at 21:12
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
@Sobi this is actually quite challenging for a beginner...
– Karn Watcharasupat
Jul 29 at 13:20
I tried to use integration by part. But it never finishes. Then when I applied the limits, it turns to pi
– Jordan Fondja
Jul 29 at 13:20
@JordanFondja this integral doesn't converge actually. If you plot the integrand out, you will notice that $e^-x^-2to1$ as $xtopminfty$ so the area under the graph till infinity can't be found.
– Karn Watcharasupat
Jul 29 at 13:23
@KarnWatcharasupat I am well aware of that. Just wanted to see how he thinks.
– Sobi
Jul 29 at 13:23
2
@KarnWatcharasupat No. It is x To the power -2
– Jordan Fondja
Jul 29 at 13:42
 |Â
show 3 more comments
@Sobi this is actually quite challenging for a beginner...
– Karn Watcharasupat
Jul 29 at 13:20
I tried to use integration by part. But it never finishes. Then when I applied the limits, it turns to pi
– Jordan Fondja
Jul 29 at 13:20
@JordanFondja this integral doesn't converge actually. If you plot the integrand out, you will notice that $e^-x^-2to1$ as $xtopminfty$ so the area under the graph till infinity can't be found.
– Karn Watcharasupat
Jul 29 at 13:23
@KarnWatcharasupat I am well aware of that. Just wanted to see how he thinks.
– Sobi
Jul 29 at 13:23
2
@KarnWatcharasupat No. It is x To the power -2
– Jordan Fondja
Jul 29 at 13:42
@Sobi this is actually quite challenging for a beginner...
– Karn Watcharasupat
Jul 29 at 13:20
@Sobi this is actually quite challenging for a beginner...
– Karn Watcharasupat
Jul 29 at 13:20
I tried to use integration by part. But it never finishes. Then when I applied the limits, it turns to pi
– Jordan Fondja
Jul 29 at 13:20
I tried to use integration by part. But it never finishes. Then when I applied the limits, it turns to pi
– Jordan Fondja
Jul 29 at 13:20
@JordanFondja this integral doesn't converge actually. If you plot the integrand out, you will notice that $e^-x^-2to1$ as $xtopminfty$ so the area under the graph till infinity can't be found.
– Karn Watcharasupat
Jul 29 at 13:23
@JordanFondja this integral doesn't converge actually. If you plot the integrand out, you will notice that $e^-x^-2to1$ as $xtopminfty$ so the area under the graph till infinity can't be found.
– Karn Watcharasupat
Jul 29 at 13:23
@KarnWatcharasupat I am well aware of that. Just wanted to see how he thinks.
– Sobi
Jul 29 at 13:23
@KarnWatcharasupat I am well aware of that. Just wanted to see how he thinks.
– Sobi
Jul 29 at 13:23
2
2
@KarnWatcharasupat No. It is x To the power -2
– Jordan Fondja
Jul 29 at 13:42
@KarnWatcharasupat No. It is x To the power -2
– Jordan Fondja
Jul 29 at 13:42
 |Â
show 3 more comments
1 Answer
1
active
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2
down vote
You probably want $-x^2$, and not $-x^-2$, because in the latter case the integral does not converge.
There is no known way to calculate this integral directly. The indirect method involves moving one dimension upwards, and looking at the double integral
$$int_-infty^inftyint_-infty^inftye^-x^2-y^2,dx,dy$$
Using some non-trivial theory, it is possible to show that this integral is precisely the square of the integral you are looking for, on one hand, and on the other hand, using some non-trivial theory of coordinate-transformations, it is possible to calculate this double integral directly, by showing that it is equal to the integral
$$2piint_0^inftyre^-r^2,dr$$
which can actually be easily computed, as the integrand is the derivative of $-e^-r^2/2$.
But you need to know a few things in order to get there. Don't waste time on trying to compute it directly. It is probably impossible.
answered Jul 29 at 13:27
uniquesolution
7,536721
I'm loss. As you said i've to know few things before I reach there.
– Jordan Fondja
Jul 29 at 13:37
@JordanFondja yeah it's really a very difficult integral to handle. You can google 'Gaussian integrals'.
– Karn Watcharasupat
Jul 29 at 13:38
@KarnWatcharasupat Ok. Thanks
– Jordan Fondja
Jul 29 at 13:40
2
From what author said, it's $e^x^-2$, not $e^-x^2$...
– Rumpelstiltskin
Jul 29 at 14:07
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You probably want $-x^2$, and not $-x^-2$, because in the latter case the integral does not converge.
There is no known way to calculate this integral directly. The indirect method involves moving one dimension upwards, and looking at the double integral
$$int_-infty^inftyint_-infty^inftye^-x^2-y^2,dx,dy$$
Using some non-trivial theory, it is possible to show that this integral is precisely the square of the integral you are looking for, on one hand, and on the other hand, using some non-trivial theory of coordinate-transformations, it is possible to calculate this double integral directly, by showing that it is equal to the integral
$$2piint_0^inftyre^-r^2,dr$$
which can actually be easily computed, as the integrand is the derivative of $-e^-r^2/2$.
But you need to know a few things in order to get there. Don't waste time on trying to compute it directly. It is probably impossible.
answered Jul 29 at 13:27
uniquesolution
7,536721
I'm loss. As you said i've to know few things before I reach there.
– Jordan Fondja
Jul 29 at 13:37
@JordanFondja yeah it's really a very difficult integral to handle. You can google 'Gaussian integrals'.
– Karn Watcharasupat
Jul 29 at 13:38
@KarnWatcharasupat Ok. Thanks
– Jordan Fondja
Jul 29 at 13:40
2
From what author said, it's $e^x^-2$, not $e^-x^2$...
– Rumpelstiltskin
Jul 29 at 14:07
add a comment |Â
up vote
2
down vote
You probably want $-x^2$, and not $-x^-2$, because in the latter case the integral does not converge.
There is no known way to calculate this integral directly. The indirect method involves moving one dimension upwards, and looking at the double integral
$$int_-infty^inftyint_-infty^inftye^-x^2-y^2,dx,dy$$
Using some non-trivial theory, it is possible to show that this integral is precisely the square of the integral you are looking for, on one hand, and on the other hand, using some non-trivial theory of coordinate-transformations, it is possible to calculate this double integral directly, by showing that it is equal to the integral
$$2piint_0^inftyre^-r^2,dr$$
which can actually be easily computed, as the integrand is the derivative of $-e^-r^2/2$.
But you need to know a few things in order to get there. Don't waste time on trying to compute it directly. It is probably impossible.
answered Jul 29 at 13:27
uniquesolution
7,536721
I'm loss. As you said i've to know few things before I reach there.
– Jordan Fondja
Jul 29 at 13:37
@JordanFondja yeah it's really a very difficult integral to handle. You can google 'Gaussian integrals'.
– Karn Watcharasupat
Jul 29 at 13:38
@KarnWatcharasupat Ok. Thanks
– Jordan Fondja
Jul 29 at 13:40
2
From what author said, it's $e^x^-2$, not $e^-x^2$...
– Rumpelstiltskin
Jul 29 at 14:07
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You probably want $-x^2$, and not $-x^-2$, because in the latter case the integral does not converge.
There is no known way to calculate this integral directly. The indirect method involves moving one dimension upwards, and looking at the double integral
$$int_-infty^inftyint_-infty^inftye^-x^2-y^2,dx,dy$$
Using some non-trivial theory, it is possible to show that this integral is precisely the square of the integral you are looking for, on one hand, and on the other hand, using some non-trivial theory of coordinate-transformations, it is possible to calculate this double integral directly, by showing that it is equal to the integral
$$2piint_0^inftyre^-r^2,dr$$
which can actually be easily computed, as the integrand is the derivative of $-e^-r^2/2$.
But you need to know a few things in order to get there. Don't waste time on trying to compute it directly. It is probably impossible.
answered Jul 29 at 13:27
uniquesolution
7,536721
You probably want $-x^2$, and not $-x^-2$, because in the latter case the integral does not converge.
There is no known way to calculate this integral directly. The indirect method involves moving one dimension upwards, and looking at the double integral
$$int_-infty^inftyint_-infty^inftye^-x^2-y^2,dx,dy$$
Using some non-trivial theory, it is possible to show that this integral is precisely the square of the integral you are looking for, on one hand, and on the other hand, using some non-trivial theory of coordinate-transformations, it is possible to calculate this double integral directly, by showing that it is equal to the integral
$$2piint_0^inftyre^-r^2,dr$$
which can actually be easily computed, as the integrand is the derivative of $-e^-r^2/2$.
But you need to know a few things in order to get there. Don't waste time on trying to compute it directly. It is probably impossible.
answered Jul 29 at 13:27
uniquesolution
7,536721
answered Jul 29 at 13:27
uniquesolution
7,536721
answered Jul 29 at 13:27
uniquesolution
7,536721
answered Jul 29 at 13:27
uniquesolution
7,536721
7,536721
I'm loss. As you said i've to know few things before I reach there.
– Jordan Fondja
Jul 29 at 13:37
@JordanFondja yeah it's really a very difficult integral to handle. You can google 'Gaussian integrals'.
– Karn Watcharasupat
Jul 29 at 13:38
@KarnWatcharasupat Ok. Thanks
– Jordan Fondja
Jul 29 at 13:40
2
From what author said, it's $e^x^-2$, not $e^-x^2$...
– Rumpelstiltskin
Jul 29 at 14:07
add a comment |Â
I'm loss. As you said i've to know few things before I reach there.
– Jordan Fondja
Jul 29 at 13:37
@JordanFondja yeah it's really a very difficult integral to handle. You can google 'Gaussian integrals'.
– Karn Watcharasupat
Jul 29 at 13:38
@KarnWatcharasupat Ok. Thanks
– Jordan Fondja
Jul 29 at 13:40
2
From what author said, it's $e^x^-2$, not $e^-x^2$...
– Rumpelstiltskin
Jul 29 at 14:07
I'm loss. As you said i've to know few things before I reach there.
– Jordan Fondja
Jul 29 at 13:37
I'm loss. As you said i've to know few things before I reach there.
– Jordan Fondja
Jul 29 at 13:37
@JordanFondja yeah it's really a very difficult integral to handle. You can google 'Gaussian integrals'.
– Karn Watcharasupat
Jul 29 at 13:38
@JordanFondja yeah it's really a very difficult integral to handle. You can google 'Gaussian integrals'.
– Karn Watcharasupat
Jul 29 at 13:38
@KarnWatcharasupat Ok. Thanks
– Jordan Fondja
Jul 29 at 13:40
@KarnWatcharasupat Ok. Thanks
– Jordan Fondja
Jul 29 at 13:40
2
2
From what author said, it's $e^x^-2$, not $e^-x^2$...
– Rumpelstiltskin
Jul 29 at 14:07
From what author said, it's $e^x^-2$, not $e^-x^2$...
– Rumpelstiltskin
Jul 29 at 14:07
add a comment |Â
@Sobi this is actually quite challenging for a beginner...
– Karn Watcharasupat
Jul 29 at 13:20
I tried to use integration by part. But it never finishes. Then when I applied the limits, it turns to pi
– Jordan Fondja
Jul 29 at 13:20
@JordanFondja this integral doesn't converge actually. If you plot the integrand out, you will notice that $e^-x^-2to1$ as $xtopminfty$ so the area under the graph till infinity can't be found.
– Karn Watcharasupat
Jul 29 at 13:23
@KarnWatcharasupat I am well aware of that. Just wanted to see how he thinks.
– Sobi
Jul 29 at 13:23
2
@KarnWatcharasupat No. It is x To the power -2
– Jordan Fondja
Jul 29 at 13:42