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My question is:

How to prove that the function:

$$f(a,b)=\int_0^\infty e^{-ax^3-bx^2}\mathrm dx$$

is a solution of the differential equation:

$$3ab\frac{{{\partial ^2}f}}{{\partial {b^2}}} - 3a\frac{{\partial f}}{{\partial b}} - 2{b^2}\frac{{\partial f}}{{\partial a}} = 1 ?$$

I have reviewed in several sites, but still nothing. :(

Thanks in advance :)

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What do you mean "I have reviewed in several places"? – t.b. Apr 20 '11 at 9:30
You could differentiate under the integral sign, you know. – J. M. Apr 20 '11 at 9:30
@J.M I think OPs real problem is evaluating either $\int e^{-ax^3-bx^2}$ or $\int x^2 e^{-ax^3-bx^2}$ over the given range. I do not think the answer will be as trivial as you indicate. – Please Delete Account Apr 20 '11 at 10:52
@Approximist: I see nothing about numerical evaluation in his post, which I'll agree is a different can of worms. Figuring out a closed form would also be interesting... – J. M. Apr 20 '11 at 11:02
@Approximist: And upgraded the question. As I said, I have differentiated $ f(a,b) = \int_0^\infty {{e^{-a{x^3}- b{x^2}}}dx}$ under the integral, obtaining $\frac{{\partial f}}{{\partial b}} = \int_0^\infty { - {x^2}{e^{ - a{x^3} - b{x^2}}}d} x$ , but becomes more complicated at the time to show that the improper integral in the variables "a" and "b" is a solution of the differential equation: $$3ab\frac{{{\partial ^2}f}}{{\partial {b^2}}} - 3a\frac{{\partial f}}{{\partial b}} - 2{b^2}\frac{{\partial f}}{{\partial a}} = 1$$ – mathsalomon Apr 20 '11 at 11:43
up vote 8 down vote accepted

If the goal of the question is to prove that $$f(a,b) = \int_0^\infty e^{-ax^3-bx^2}\mathrm dx,$$ satisfies $$3ab\frac{\partial^2 f}{\partial b^2}-3a\frac{\partial f}{\partial b}-2b^2\frac{\partial f}{\partial a} = 1,$$ then this can be done by looking at all terms at the same time, not separately.

I will assume that the differentiating under the integral sign is not the problematic part.

By plugging in, we get the following: $$3ab\frac{\partial^2f}{\partial b^2}-3a\frac{\partial f}{\partial b}-2b^2\frac{\partial f}{\partial a} = \int_0^\infty \left(3abx^4+3ax^2+2b^2 x^3\right)e^{-ax^3-bx^2}\mathrm dx.$$

Now, observe that $$\int_0^\infty 3ax^2e^{-ax^3 - bx^2}\mathrm dx = \int_0^\infty ax^3\left(3ax^2 + 2bx\right)e^{-ax^3-bx^2}\mathrm dx ,$$ by integration by parts.

This means that, after factoring, we obtain $$\int_0^\infty \left(3abx^4+3ax^2+2b^2 x^3\right)e^{-ax^3-bx^2}\mathrm dx = \int_0^\infty \left(3ax^2+2bx\right) \left(ax^3+bx^2\right) e^{-ax^3-bx^2}\mathrm dx.$$ Using the substitution $u = ax^3 + bx^2$, this reduces to $$\int_0^\infty ue^{-u}\mathrm du = 1,$$ whence we get the wanted identity.

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Nice work! $\;$ – J. M. Apr 20 '11 at 14:28
Nice. By the way, this is a good example that shows that it's good to include context in the question and say what one is more broadly interested in, rather than isolating the part of it that one is currently focussing on. – joriki Apr 20 '11 at 14:57
@J. M. Thank you very much, it became very clear reply. Now I know how to derive this type of integral functions :) – mathsalomon Apr 20 '11 at 15:38
@mathsalomon: Thank Raeder, not me. I merely formatted his/her post. – J. M. Apr 20 '11 at 15:40
@mathsalomon : You're welcome! It would be great if you go edit the question so as to reflect what you were really asking :) – Raeder Apr 20 '11 at 16:12

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