Good morning,
I had this demonstration to do on my calculus exam.
$$I_n=\int x^ne^{ax^2}$$
This integration is equal to: $$I_n=\frac{1}{2a}x^{n-1}e^{ax^2}-\frac{n-1}{2a} I_{n-2}.$$
Please help me.
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$\begingroup$ Please use Mathjax for typing your question. $\endgroup$– NizarApr 8, 2016 at 8:32
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1$\begingroup$ What is the question? Are you asking how to derive that expression for $I_n$ in terms of $I_{n-2}$? $\endgroup$– almagestApr 8, 2016 at 8:39
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$\begingroup$ @almagest I think he meant $ I_n = \displaystyle\int x^n e^{ax^2} \mathrm{d}x $ . $\endgroup$– LouaApr 8, 2016 at 10:35
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$\begingroup$ If it's for an exam, ask the professor. $\endgroup$– anomalyApr 9, 2016 at 16:16
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1 Answer
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If $$I_n=\int x^ne^{ax^2}dx$$ use integration by parts $$u=x^{n-1}\implies u'=(n-1)x^{n-2}$$ $$v'=x e^{ax^2}dx\implies v=\frac{e^{a x^2}}{2 a}$$ I am sure that you can take it from here.
answered Apr 8, 2016 at 8:39
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$\begingroup$ I doubt it, since this is a really simple case of integration by parts! $\endgroup$– almagestApr 8, 2016 at 8:50
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$\begingroup$ @almagest. I agree that this is probably one of the simplest cases. Cheers. $\endgroup$ Apr 8, 2016 at 8:52