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I wonder how to find an expression for approximating: $$ f(x)=\int_{x}^\infty \frac{x}{\sigma a^2 {\sqrt {2\pi (a^2-x^2)}} } e^{-\frac{(\log(a)- \mu )^2}{2\sigma^2}}\mathrm{d}a $$

I don't know if it can be expressed in terms elementary functions exactly. In what direction or study topics do I need to approximate it?

My further aim is to deal with the variable $x$ with the probability distribution function $a(x-b)f(x)$ where $a$, $b$ are positive constants. I would then estimate $\sigma$ and $\mu$ from measurements (of some real geological objects).I am in the middle of undergraduate mathematical studies.

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    $\begingroup$ Your integral is kind of confusing. You have $x$ as both a local and global variable, but you're integrating over $a$. $\endgroup$ – Ben Dec 25 '17 at 23:37
  • $\begingroup$ I am doing exactly this, but how to do it? $\endgroup$ – Galtulling Dec 26 '17 at 0:01
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I would first make the change of variables $a=x\tau$, resulting in $$ f(x)=\int_{x}^\infty \frac{x}{\sigma a^2 {\sqrt {2\pi (a^2-x^2)}} } e^{-\frac{(\log(a)- \mu )^2}{2\sigma^2}}\mathrm{d}a= $$ $$\frac{1}{\sigma |x|\sqrt{2\pi}}\int_1^\infty\frac{\mathrm{d}\tau}{\tau^2\sqrt{\tau^2-1}}\exp\left[-\frac{(\log\tau+\log x-\mu)^2}{2\sigma^2}\right]\ . $$ Next, I would call $\log\tau=z$, resulting in $$ \frac{1}{\sigma |x|\sqrt{2\pi}}\int_0^\infty\frac{\mathrm{d}z}{e^z\sqrt{e^{2z}-1}}\exp\left[-\frac{(z+\log x-\mu)^2}{2\sigma^2}\right]\ . $$ This integral still cannot be expressed in closed form. However, a systematic approximation scheme can be constructed by a series expansion $$ \frac{1}{\sqrt{X-1}}\simeq \frac{1}{\sqrt{X}}+\frac{1}{2}\frac{1}{X^{3/2}}+\frac{3}{8}\frac{1}{X^{5/2}}+\ldots\ , $$ applied to $X=e^{2z}$. Therefore the integral (with $A=\mu-\log x$) $$ \int_0^\infty\frac{\mathrm{d}z}{e^z\sqrt{e^{2z}-1}}\exp\left[-\frac{(z-A)^2}{2\sigma^2}\right]\simeq \int_0^\infty\frac{\mathrm{d}z}{e^z\times e^z}\exp\left[-\frac{(z-A)^2}{2\sigma^2}\right]+$$ $$\frac{1}{2}\int_0^\infty\frac{\mathrm{d}z}{e^z\times e^{3z}}\exp\left[-\frac{(z-A)^2}{2\sigma^2}\right]+\frac{3}{8}\int_0^\infty\frac{\mathrm{d}z}{e^z\times e^{5z}}\exp\left[-\frac{(z-A)^2}{2\sigma^2}\right]+\ldots\ , $$ Each of these integrals can now be expressed in closed form, and the approximation can be made as good as you wish by including more and more terms. For example, including the first three terms as above, the approximate expression reads $$ f(x)\simeq \frac{1}{\sigma |x|\sqrt{2\pi}}\left[\frac{1}{8} \sqrt{\frac{\pi }{2}} \sigma e^{2 \sigma ^2-6 A} \left(8 e^{4 A} \left(\text{erf}\left(\frac{A-2 \sigma ^2}{\sqrt{2} \sigma }\right)+1\right)+4 e^{2 A+6 \sigma ^2} \left(\text{erf}\left(\frac{A-4 \sigma ^2}{\sqrt{2} \sigma }\right)+1\right)+3 e^{16 \sigma ^2} \left(\text{erf}\left(\frac{A-6 \sigma ^2}{\sqrt{2} \sigma }\right)+1\right)\right)\right]\ , $$ where $\mathrm{erf}$ is the Error function.

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  • $\begingroup$ Wow! Thank you very very much for the solution! It is great and extremely helpful! $\endgroup$ – Galtulling Dec 26 '17 at 8:51

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