Is there a closed form for $$\int_{0}^{\infty}e^{ax^3+bx^2}\,\mathrm{d}x $$?
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3no need for a closed form because the definite integral is not convergent anyways (supposing $a$ not null). – JJacquelin Apr 04 '14 at 13:15
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@OJB I tried it with WolframAlpha, and even with extra computational time it still timed out, so it's unlikely that there's a simple closed form expression. – Keshav Srinivasan Apr 04 '14 at 13:22
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I assume $b \le 0$ as well. It will be convergent but finding a closed form is something else... – user88595 Apr 04 '14 at 13:23
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@KeshavSrinivasan So? This is clearly beyond WA. I'm 99.99% sure a closed form exists for this. – user85798 Apr 04 '14 at 13:25
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$\displaystyle\int_0^\infty e^{-x^n}dx=\Gamma\Big(1+\tfrac1n\Big)$ – Lucian Apr 04 '14 at 13:27
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One can always Taylor expand $e^{bx^2}$, integrate the expression term by term and turn the integral into a sum of 3 generalized hypergeometric functions. I think it is a sum of 3 $_2F_2$. – achille hui Apr 04 '14 at 13:31
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@OJB What makes you so sure? – Keshav Srinivasan Apr 04 '14 at 13:31
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@KeshavSrinivasan I found this on wiki https://upload.wikimedia.org/math/c/c/3/cc3a9fca8785f9c913757462a05cee87.png – user85798 Apr 04 '14 at 13:33
2 Answers
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Let $a<0$, and changing variable $y = (-a x^3)$: $$ \int_0^\infty \exp(a x^3 + b x^2) \mathrm{d}x = \alpha\int_0^\infty \exp\left(-y + \beta y^{2/3} \right) y^{-2/3} \mathrm{d}y $$ where $\alpha = 1/\left(3 (-a)^{1/3}\right)$ and $\beta = b (-a)^{-2/3}$.
Now, using $y^{-2/3} \exp\left(\beta y^{2/3}\right) = \sum_{m=0}^\infty \frac{1}{m!} \beta^{m} y^{2 (m-1)/3}$, and interchanging the order summation and integration as warranted by Tonelli theorem: $$ \int_0^\infty \exp\left(-y + \beta y^{2/3} \right) y^{-2/3} \mathrm{d}y = \sum_{m=0}^\infty \frac{\beta^m}{m!} \int_0^\infty \mathrm{e}^{-y} y^{2 (m-1)/3} \mathrm{d}y = \sum_{m=0}^\infty \frac{\beta^m}{m!} \Gamma\left(\frac{2m}{3} + \frac{1}{3}\right) $$ Using duplication and triplication formulae for $\Gamma$-function $$ \Gamma(2x) = \frac{2^{2x}}{2 \sqrt{\pi}} \Gamma(x) \Gamma\left(x+\tfrac{1}{2}\right) \quad \Gamma(3x) = \frac{3^{3x}}{2 \sqrt{3} \pi} \Gamma(x) \Gamma\left(x+\tfrac{1}{3}\right) \Gamma\left(x+\tfrac{2}{3}\right) $$ we can rewrite the summand: $$ \frac{\Gamma\left(\frac{2m}{3} + \frac{1}{3}\right)}{m!} = \frac{\Gamma\left(\frac{2m}{3} + \frac{1}{3}\right)}{\Gamma(m+1)} = \frac{2^{1/3} \sqrt{\pi}}{\sqrt{3}} \left(\frac{2^{2/3}}{3}\right)^m \frac{\Gamma\left(\frac{m}{3} + \frac{1}{6}\right)}{\Gamma\left(\frac{m}{3} + 1\right) \Gamma\left(\frac{m}{3} + \frac{1}{3}\right) } $$ It now suffices to split the sum $$\sum_{m=0}^\infty g(m) = \sum_{k=0}^\infty g(3k) + \sum_{k=0}^\infty g(3k+1) + \sum_{k=0}^\infty g(3k+2) $$ and write the answer in terms of hypergeometric functions: $$ \begin{eqnarray} \int_0^\infty \exp(a x^3 + b x^2) \mathrm{d}x &=& \tilde{\alpha} \frac{\Gamma(1/6)}{\Gamma(1/3)} \cdot {}_1F_1\left(\tfrac{1}{6}; \tfrac{1}{3}; \left(2^{2/3} \frac{\beta}{3}\right)^3 \right) \\ &+& \tilde{\alpha} \cdot \frac{2^{2/3} \beta}{3} \cdot \frac{\Gamma(1/2)}{\Gamma(2/3)\Gamma(4/3)} \cdot {}_2F_2\left(1, \tfrac{1}{2}; \tfrac{2}{3},\tfrac{4}{3}; \left(2^{2/3} \frac{\beta}{3}\right)^3 \right) \\ &+& \tilde{\alpha} \cdot \left(\frac{2^{2/3} \beta}{3}\right)^2 \cdot \frac{\Gamma(5/6)}{\Gamma(1)\Gamma(5/3)} \cdot {}_1F_1\left(\tfrac{5}{6}; \tfrac{5}{3}; \left(2^{2/3} \frac{\beta}{3}\right)^3 \right) \end{eqnarray} $$ where $\tilde{\alpha} = \frac{2^{1/3} \sqrt{\pi}}{\sqrt{3}}\alpha $
Sasha
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Here is what WolframAlpha gives me, just for $a=-1$:
$$\begin{align*} \int^\infty_0 e^{-x^3+bx^2}\,dx=-\frac1{27}b\,\left(2\sqrt{3}\pi e^{\tfrac{2b^3}{27}}\left(I_{-\frac13}\left(-\dfrac{2b^3}{27}\right)+I_{\frac13}\left(-\dfrac{2b^3}{27}\right)\right)-9_2F_2\left(\dfrac{1}{2},1;\dfrac23,\dfrac43;\dfrac{4b^3}{27}\right)\right)\,\,\,\\\color{grey}{\text{ for }}\,\operatorname{Re}(b)<0 \end{align*} $$
It's expressed using Bessel functions and hypergeometric functions, and WolframAlpha was just barely able to compute it with extra computational time. So just imagine how complicated an expression you'd have in the general case.
Keshav Srinivasan
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3A simple substitution, with $x=a^{1/3}y$, and a different $b$, will give the general case – Empy2 Apr 04 '14 at 13:32