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I would like to calculate $$\frac{\partial^{1/2}}{\partial x^{1/2}}\left( e^{-\alpha x^2 + \beta x} \right) $$ My intuition is that I would have to use some sort of fractional Leibniz formula to first separate calculus of the half derivative of $e^{-\alpha x^2}$ from the other one, which is easy to derive $$ \frac{\partial^{1/2}}{\partial x^{1/2}}\left( e^{\beta x} \right) = \sqrt{\beta} e^{\beta x} $$ Thus two questions arise from there

  • Does a fractional version of Leibniz theorem exists ?
  • Does a fractional version of Hermite polynomial exists ? In other words, how could we calculate $ \frac{\partial^{1/2}}{\partial x^{1/2}}\left( e^{-\alpha x^2} \right)$ using $ \frac{\partial^{n}}{\partial x^{n}}\left( e^{-\alpha x^2} \right) = (-1)^n H_n(x)e^{-\alpha x^2}$ ?
vanna
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2 Answers2

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The Riemann-Liouville formula for the semiderivative goes like this:

$${}_0 D_x^{(1/2)} f(x)=\frac{f(0)}{\sqrt{\pi x}}+\frac1{\sqrt\pi}\int_0^x \frac{f^\prime(t)}{\sqrt{x-t}}\mathrm dt$$

Taking $f(x)=\exp(-\alpha x^2+\beta x)$, we then have

$$\begin{align*} {}_0 D_x^{(1/2)} f(x)&=\frac1{\sqrt{\pi x}}+\frac1{\sqrt\pi}\int_0^x \frac{(\beta-2\alpha t)\exp(-\alpha t^2+\beta t)}{\sqrt{x-t}}\mathrm dt\\ &=\frac1{\sqrt{\pi x}}+\frac1{\sqrt\pi}\left(\beta\int_0^x \frac{\exp(-\alpha t^2+\beta t)}{\sqrt{x-t}}\mathrm dt-2\alpha\int_0^x \frac{t\exp(-\alpha t^2+\beta t)}{\sqrt{x-t}}\mathrm dt\right) \end{align*}$$

Unfortunately, neither integral seems to have an easy evaluation in terms of known functions unless $\alpha=0$ or $\beta=0$, in which case,

$$\begin{align*} {}_0 D_x^{(1/2)} \exp(-\alpha x^2)&=\frac1{\sqrt{\pi x}}-\frac{8\alpha x^{3/2}}{3\sqrt\pi}{}_2 F_2\left({{1,\frac32}\atop{\frac54,\frac74}}\mid -\alpha x^2\right)\\ {}_0 D_x^{(1/2)} \exp(\beta x)&=\frac1{\sqrt{\pi x}}+\sqrt{\beta}\exp(\beta x)\mathrm{erf}(\sqrt{\beta x}) \end{align*}$$

There is in fact a semiderivative version of the Leibniz formula, but as I said, it looks somewhat unwieldy:

$${}_0 D_x^{(1/2)}f(x)g(x)=\sum_{j=0}^\infty \binom{\frac12}{j}\left({}_0 D_x^{(1/2-j)}f(x)\right)g^{(j)}(x)$$

If we take $f(x)=\exp(\beta x)$ and $g(x)=\exp(-\alpha x^2)$, we get

$$\begin{split}&{}_0 D_x^{(1/2)} \exp(-\alpha x^2+\beta x)=\\&\quad\exp(-\alpha x^2+\beta x)\sum_{j=0}^\infty \binom{\frac12}{j}\frac{(-1)^j \alpha^{j/2} \beta^{1/2-j}}{\Gamma(j-1/2)}\gamma(j-1/2,\beta x) H_j(x\sqrt\alpha)\end{split}$$

where $\gamma(a,x)$ is a (lower) incomplete gamma function, and $H_j(x)$ is a Hermite polynomial. I know of no simpler way to express this infinite sum, however.

J. M. ain't a mathematician
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  • You're welcome. I'm sorry I can't be more helpful... – J. M. ain't a mathematician Jul 13 '12 at 13:43
  • It is pretty much what I got so far yes, but it is well-synthesized. The Riemann-Liouville fractional integral is not giving exactly the results I was expecting. I would like to find a way to differentiate $e^{-\alpha x}$ w.r.t. $\alpha$ such that it yields $\sqrt{x}e^{-\alpha x}$ (I actually want to use it under the integration sign to differentiate probability characteristic functions). Do you think I should look at other type of fractional calculus (Caputo e.g.) or is it a dead end ? – vanna Jul 13 '12 at 13:51
  • When I tried it out, the Caputo form of the semiderivative gave the same results, so it doesn't seem to be any more useful than what I've shown. – J. M. ain't a mathematician Jul 13 '12 at 13:53
  • In fact I just want a theory of differentiation that would commute with the Lebesgue integral under classic regularity conditions. What about this http://en.wikipedia.org/wiki/Gr%C3%BCnwald%E2%80%93Letnikov_derivative ? It may lead to different results. – vanna Jul 13 '12 at 13:57
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    @vanna:, well, the Grünwald–Letnikov definition can be shown to be equivalent to the Riemann-Liouville definition (see here for a proof), and it's a bit more unwieldy to use in practice... – J. M. ain't a mathematician Jul 13 '12 at 14:08
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J. M. 's answer is very detailed and nice. I just wanted to add another approach for fractial derivatives by using power series method .

we know from wikipedia that $$\frac{\partial^{n}}{\partial x^{n}}(x^m)= \frac{\Gamma{(m+1)}}{\Gamma{(m-n+1)}} x^{m-n}$$

$$\frac{\partial^{n}}{\partial x^{n}}((x+a)^m)= \frac{\Gamma{(m+1)}}{\Gamma{(m-n+1)}} (x+a)^{m-n}$$

$$\frac{\partial^{1/2}}{\partial x^{1/2}}\left( e^{-\alpha x^2 + \beta x} \right) =\frac{\partial^{1/2}}{\partial x^{1/2}}\left( e^{\alpha a^2} e^{-\alpha (x +a)^2 } \right)$$

$-2 \alpha a = \beta $

thus $ a = -\frac{\beta}{2 \alpha} $

$\frac{\partial^{1/2}}{\partial x^{1/2}}\left( e^{\alpha a^2} e^{-\alpha (x +a)^2 } \right)= e^{\alpha a^2} \frac{\partial^{1/2}}{\partial x^{1/2}}(\sum \limits_{k=0}^\infty \frac{(-\alpha)^k(x+a)^{2k}}{\Gamma{(k+1)}})=$

$=e^{\alpha a^2} \sum \limits_{k=0}^\infty \frac{(-\alpha)^k}{\Gamma{(k+1)}}\frac{\partial^{1/2}}{\partial x^{1/2}}( (x+a)^{2k})=e^{\alpha a^2} \sum \limits_{k=0}^\infty \frac{(-\alpha)^k \Gamma{(2k+1)}(x+a)^{2k-\frac{1}{2}}}{\Gamma{(k+1)} \Gamma{(2k+\frac{1}{2})}}=$

$=\frac{e^{\alpha a^2}}{\sqrt{(x+a)}} \sum \limits_{k=0}^\infty \frac{(-\alpha)^k \Gamma{(2k+1)}(x+a)^{2k}}{\Gamma{(k+1)} \Gamma{(2k+\frac{1}{2})}}$

$$\frac{\partial^{1/2}}{\partial x^{1/2}}\left( e^{-\alpha x^2 + \beta x} \right)=\frac{e^{ \frac{\beta^2}{4 \alpha}}}{\sqrt{x-\frac{\beta}{2 \alpha}}} \sum \limits_{k=0}^\infty \frac{(-\alpha)^k \Gamma{(2k+1)}(x-\frac{\beta}{2 \alpha})^{2k}}{\Gamma{(k+1)} \Gamma{(2k+\frac{1}{2})}}$$

Note: If I simplfy the result more I will edit.

Mathlover
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