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.
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@Sobi this is actually quite challenging for a beginner... – Karn Watcharasupat Jul 29 '18 at 13:20
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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 '18 at 13:20
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@JordanFondja this integral doesn't converge actually. If you plot the integrand out, you will notice that $e^{-x^{-2}}\to1$ as $x\to\pm\infty$ so the area under the graph till infinity can't be found. – Karn Watcharasupat Jul 29 '18 at 13:23
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@KarnWatcharasupat I am well aware of that. Just wanted to see how he thinks. – MSDG Jul 29 '18 at 13:23
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@Sobi actually without putting the limits, the integration never finishes. Is there another method apart from bypart To solve it? – Jordan Fondja Jul 29 '18 at 13:26
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@KarnWatcharasupat ok. Now i have change the question. From -x To x. Can this converge? – Jordan Fondja Jul 29 '18 at 13:36
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By the way, do you mean $e^{-x^2}$ for the integrand though? If yes, check this link: https://en.wikipedia.org/wiki/Gaussian_integral#By_Cartesian_coordinates – Karn Watcharasupat Jul 29 '18 at 13:39
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2@KarnWatcharasupat No. It is x To the power -2 – Jordan Fondja Jul 29 '18 at 13:42
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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
$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-x^2-y^2}\,dx\,dy$$
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 $$2\pi\int_0^{\infty}r{e^{-r^2}}\,dr$$ 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.
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I'm loss. As you said i've to know few things before I reach there. – Jordan Fondja Jul 29 '18 at 13:37
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@JordanFondja yeah it's really a very difficult integral to handle. You can google 'Gaussian integrals'. – Karn Watcharasupat Jul 29 '18 at 13:38
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