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While attempting to teach myself the fractional calculus, I encountered a tragically early roadblock. For non-power rule fractional derivatives, I am having a lot of trouble evaluating for a closed form.

Would someone mind walking me through the process for taking the half-derivative of $$f(x) = e^x$$

Really the most difficult part is evaluating

$$\int_0^x \frac{e^t}{\sqrt{x-t}} dt$$

but a full hand-holding would be really helpful.

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    $\begingroup$ I don't think you've written down the integral you actually need help with. At the very least, it should probably have an $e^t$ in the integrand rather than an $e^x$. $\endgroup$ – Barry Cipra Jun 12 '15 at 17:03
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    $\begingroup$ Shouldn't the integral involve $\sqrt{x - t}$ rather than $x - t$? $\endgroup$ – anomaly Jun 12 '15 at 17:05
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    $\begingroup$ Are you 100% sure you have the right integral now? Because had almost wriiten down the second answer with again the wrong integral. $\endgroup$ – wythagoras Jun 12 '15 at 17:07
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    $\begingroup$ Don't worry about it. $\endgroup$ – wythagoras Jun 12 '15 at 17:09
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    $\begingroup$ @craft94, you can do "penance" by upvoting the correct answers to the original version of the question, even though they're no longer relevant. $\endgroup$ – Barry Cipra Jun 12 '15 at 17:12
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For the integral: Keep in mind that $x$ is a constant!

$$\int_0^x \frac{e^t}{\sqrt{x-t}} dt$$

Use the substitution $u=x-t$, then $du=-dt$. This gives:

$$\int_0^x -\frac{e^{x-u}}{\sqrt{u}} du$$

$$-e^x\int_0^x \frac{e^{-u}}{\sqrt{u}} du$$

$$-e^x\int_0^x u^{-1/2}e^{-u} du$$

$$-e^x\gamma\left(\frac{1}{2},x\right)$$

Where $\gamma$ is the incomplete lower gamma function.

This can also be written as $$-e^x \sqrt{x} E_{\frac{1}{2}}(x)$$

using the exponential integral function. It has been proven there is no closed form of this function.

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  • $\begingroup$ So does that mean that the half-derivative doesn't exist? Or is infinite? I really wish there more intuition to be garnered from fractional-calculus. $\endgroup$ – craft94 Jun 12 '15 at 17:13
  • $\begingroup$ @craft94 It means that there is no closed form. $\endgroup$ – user223391 Jul 23 '15 at 2:29
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Let we assume that $x>0$. Since: $$ e^x = \sum_{n\geq 0}\frac{x^n}{n!}\tag{1} $$ and: $$ D^{1/2} x^{m} = \frac{x^{m-1/2}\,\Gamma\!\left(m+1\right)}{\Gamma\!\left(m+\frac{1}{2}\right)}\tag{2}$$ (look here, for instance) we have: $$ D^{1/2} e^x = \frac{1}{\sqrt{x}}\sum_{n\geq 0}\frac{x^n}{\Gamma\!\left(n+\frac{1}{2}\right)}=\frac{1+e^x\sqrt{\pi x}\;\text{Erf}(\sqrt{x})}{\sqrt{\pi x}}\tag{3}$$ where $\text{Erf}$ is the usual error function.

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$${1 \over {\Gamma(1/2)}} \cdot {{d} \over {dx}} \int_0^x {{e^t} \over {\sqrt {x-t}}} \ dt$$ Where $\Gamma(x)$ is the generalized factorial function. This equals $${1 \over {\Gamma(1/2)}} \cdot {{d} \over {dx}} \int_0^x {{e^t} \over {\sqrt {x-t}}} \ dt=e^x \cdot \operatorname{erf}(\sqrt{x})$$ where $\operatorname{erf}(u)$ is the error function. This is more of a definition than a technical thing, so you don't really need to prove the above per se. However the substitution $u=\sqrt{x-t}$ and integration by parts brings the above into compliance with $$\operatorname{erf}(x)={2 \over {\sqrt{\pi}}} \cdot \int_0^x e^{-t^2} \ dt$$

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If this is really the integral you want, then notice you are integrating with respect to $t$ so treat $x$ as a constant. Then $$\int \frac{e^x}{x-t} dt = e^x\int\frac{1}{x-t} dt$$ and the antiderivative of $\frac{1}{x-t}$ with respect to $t$ is easily seen to be $-\ln(x-t)+C$. Now evaluate the integral as you normally would with a definite integral.

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This depends on a method and definition used, but all the other answers give unnatural expressions in my view.

The following definitions give a more natural answer and coincide with each other.

The first one is based on Newton series interpolation over consecutive integer derivatives:

$$f^{(s)}(x)=\sum_{m=0}^{\infty} \binom {s}m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$

the other one is based on Forier transform:

$$f^{(s)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} e^{- i \omega x}{\omega}^s \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

For $f(x)=e^x$ they both give the same result: the derivative of any order of this function is $e^x$. In other words, this function is invariant regarding differentiation and integration of any order.

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