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I recently came across how to $\int e^{-x^2}dx$ by "change of variable". i.e.,

let $I = \int e^{-x^2}dx$
then, $I^2 = \int e^{-x^2}dx\int e^{-x^2}dx$
and changing $x$ to $y$ in one of the integrals above to make it $I^2 = \int e^{-x^2}dx\int e^{-y^2}dy$ and then convert it to polar coordinates, find the integral and take the squate root of the integral.

I think i understand this very well, what I wanted to know is that if I can use the same technique to evaluate $\int \sin(x)e^{-x^2}dx$. I would appreciate if anyone could shed some light on this. Thanks.

S L
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Yash
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    That technique only works to find the definite integral over the entire real line (or half of it). It does not actually yield an indefinite integral for $e^{-x^2}$ in terms of elementary functions. –  May 11 '14 at 20:00
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    The same technique can be used if the question is $$ \int_0^\infty\cos x\ e^{\Large -x^2}\ dx. $$ – Tunk-Fey May 12 '14 at 09:49

1 Answers1

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This is not going to work, since you do not have $\sin x \sin y = \sin xy$, hence the integrand does not factor when expressed in polar coordinates. Instead, you can calculate this integral is either by complex analysis or by differentiation under the integral sign. (Assuming you mean the integral from $0$ to $\infty$).

user111187
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