My classmate asked me about this integral:$$\int_0^{\infty} e^{-x^3+2x^2+1}\,\mathrm{d}x$$ but I have no idea how to do it. What's the closed form of it? I guess it may be related to the Airy function.
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Renascence_5.
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According to Liouville's theorem on primitives, there is no expression in terms of simple functions. – Tom-Tom Jan 05 '16 at 10:02
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Would the answer in this thread help? Tom-Tom is right, but maybe you can write it in terms of power series. – Eric S. Jan 05 '16 at 10:03
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1Is your classmate a friend ? Because such a gift would be making me crazy ! By the way, your guess about Airy function was very good. Cheers. – Claude Leibovici Jan 05 '16 at 10:07
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1@ClaudeLeibovici I will kill him- -! – Renascence_5. Jan 05 '16 at 10:09
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Try to find another one to challenge him ! There are plenty of them in this splendid site. – Claude Leibovici Jan 05 '16 at 10:11
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@Tom-Tom Special function？ – Renascence_5. Jan 05 '16 at 10:18
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2I've tried to define $$\varphi(\lambda)=\int_0^\infty \exp\left[-x^3-3\lambda x^2-3\left(\lambda^2-\frac43\right)x-\lambda^3+\frac43 \lambda-\frac{37}{27}\right]\mathrm dx. $$You want $\varphi(-\frac23)$. Computing $\varphi'(\lambda)$ gives $\varphi'(\lambda)=\frac{37}{27}\varphi(\lambda)+\exp(-\lambda^3+\frac43\lambda-\frac{37}{27})$ but I can't see how to go beyond that. – Tom-Tom Jan 05 '16 at 10:24
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I guess it may be related to the Airy function.
You guessed well. In general, we have $~\displaystyle\int_0^\infty\exp\Big(-x^2(x+3a)\Big)~dx~=~\frac2{e^{2a^3}}\cdot\frac{\text{Bi}\Big(3^{2/3}~a^2\Big)}{3^{4/3}}-$
$-a\cdot~_2F_2\bigg(\bigg[\dfrac12~,~1\bigg]~;~\bigg[\dfrac23~,~\dfrac43\bigg]~;~-4a^3\bigg).~$ In this particular case, $a=-\dfrac23.$
Lucian
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Apparently there is a closed form that includes the generalized hypergeometric functions. They themselves can be expressed in a series, so you might be able to trace back the solution to some kind of series expression of the integrand:
http://www.wolframalpha.com/input/?i=Integrate+Exp[-x^3%2B2+x^2+%2B1]+from+0+to+Infinity
My approach would be to take the solution and write the functions as a hypergeometric series.
Asking Questions
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I also don't know how helpful that approach will actually be...but for the record here are the relations: http://mathworld.wolfram.com/GeneralizedHypergeometricFunction.html and http://mathworld.wolfram.com/AiryFunctions.html – Asking Questions Jan 05 '16 at 10:43