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Everyone, I am try to evaluate this integral for one day, could anyone give me some hints and help me solve this integral? $$ \int_0^{\infty}\frac{r}{1+s^{-1}[r^{-\alpha}+{(r+d)}^{-\alpha}]^{-1}}dr $$ $s$, $\alpha$, and $d$ are all constant, and $\alpha > 2$. Thank you very much!

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  • $\begingroup$ Hi! what have you tried so far? $\endgroup$ – MattAllegro May 20 '14 at 19:35
  • $\begingroup$ Hi! First I give some values of $s$, $\alpha$, and $d$, and use MATLAB to calculate this integral, it can give me an answer. So I believe this integral can be calculated. Then, I try to replace the %r^{-\alpha}% with $x$, but it seems make it more difficult. I don not have good idea to solve this problem. Could you give me suggestion? Thank you! $\endgroup$ – BlueSky May 20 '14 at 19:43
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Hint: the integral you have defined depends on three real parameters:

$$I(\alpha,s,d):=\int_0^{\infty}\frac{r}{1+s^{-1}[r^{-\alpha}+{(r+d)}^{-\alpha}]^{-1}}\mathbb{d}r.$$

When dealing with complex multi-parameter integrals like these, a smart first step is to rescale the independent variable since this almost always allows you to reduce the number of parameters in the integral by one.

Substitute $r=d\tilde{r}$:

$$\frac{r}{1+s^{-1}[r^{-\alpha}+{(r+d)}^{-\alpha}]^{-1}} = \frac{d\,\tilde{r}}{1+s^{-1}[d^{-\alpha}\tilde{r}^{-\alpha}+{(d\tilde{r}+d)}^{-\alpha}]^{-1}}\\ = \frac{d\,\tilde{r}}{1+s^{-1}[d^{-\alpha}\tilde{r}^{-\alpha}+{d^{-\alpha}(\tilde{r}+1)}^{-\alpha}]^{-1}}\\ = \frac{d\,\tilde{r}}{1+s^{-1}d^{\alpha}[\tilde{r}^{-\alpha}+{(\tilde{r}+1)}^{-\alpha}]^{-1}}.$$

Then,

$$I(\alpha,s,d)=d^2\int_{0}^{\infty}\frac{\tilde{r}}{1+s^{-1}d^{\alpha}[\tilde{r}^{-\alpha}+{(\tilde{r}+1)}^{-\alpha}]^{-1}}\mathbb{d}\tilde{r}\\ =d^2I(\alpha,\frac{s}{d^{\alpha}},1).$$

This reduces the problem to evaluating the two parameter integral,

$$I(\alpha,s)=\int_0^{\infty}\frac{r}{1+s^{-1}[r^{-\alpha}+{(r+1)}^{-\alpha}]^{-1}}\mathbb{d}r.$$

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  • $\begingroup$ Thank you very much. Very Nice Substitution, and smart idea. Thanks again! $\endgroup$ – BlueSky May 20 '14 at 19:59
  • $\begingroup$ Even for this two parameter integral, it still seems hard to calculate. Do you have some suggestion? $\endgroup$ – BlueSky May 20 '14 at 20:17
  • $\begingroup$ @BlueSky Unfortunately, no. I'm still toying around with it, but at the moment I really don't know what else to tell you. I hope you don't mind that I posted an incomplete answer. I just figured you'd prefer a partial answer to no answer. :) $\endgroup$ – David H May 20 '14 at 20:38
  • $\begingroup$ Hi,David. I do appreciate you give me your smart thinking, even it is an incomplete answer. I agree this is very important step which can reduce one parameter. Thanks again! :) $\endgroup$ – BlueSky May 20 '14 at 20:52
  • $\begingroup$ Hi, David. Inspired by your idea, I try to substitute $s^{-\frac{1}{\alpha}}r$ with $x$. Will this substitution help? $\endgroup$ – BlueSky May 20 '14 at 21:17

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