Closed form for the following integral

Question

How can I find the closed form for the following please?

$$f(r) = \frac{1}{R_x+R_C}\int^{R_x+R_c}_{0}\frac{1}{1+sr^{-4}} dr$$

where $R_x,R_c$ are constants

Answer 1

A couple things: the integral doesn't depend on $r$; i.e. you won't get out a function $f(r)$ because the $r$ is integrated out. Also, there is (possibly) a singularity in the range of integration if $s < 0$ so I'll assume that $s > 0$ (if $s = 0$ then the integral is trivial).

Put $T = R_x + R_c$. Then the integral is \begin{align*} I &= \frac{1}{T}\int^T_0 \frac{dr}{1+sr^{-4}} \\&= \frac{1}{T}\int^T_0 \frac{r^4 dr}{r^4 + s} \\&= \frac{1}{T} \int^T_0 \left(\frac{r^4+s}{r^4 + s} - \frac{s}{r^4 + s} \right) dr \\&= 1 - \frac{s}{T} \int^T_0\frac{dr}{r^4 + s}. \end{align*} Let $r = s^{1/4}x$. Then $$I = 1 - \frac{s}{T} \int^{s^{-1/4}T}_0 \frac{s^{1/4}dx}{s(x^4+1)} = 1- \frac{1}{S}\underbrace{\int^{S}_0 \frac{dx}{x^4 +1}}_{\,\,\,\, = I^*}$$ where $S = s^{-1/4}T$. We focus on evaluating $I^*$. Now \begin{align*}\frac{1}{x^4+1} &= \frac{1}{x^4 + 2x^2 +1 - 2x^2} \\ &= \frac{1}{(x^2+1)^2 - 2x^2} \\&= \frac{1}{(x^2 +1 -x\sqrt 2)(x^2+1 + x\sqrt 2)} \\ &= \frac{1}{2\sqrt 2} \left(\frac{\sqrt 2 + x}{x^2 + x\sqrt 2 + 1} + \frac{\sqrt 2 - x}{x^2 - x\sqrt 2 + 1}\right) \\ &= \frac{1}{2\sqrt 2} \left(\frac{\sqrt 2 + x}{(x+\sqrt2/2)^2 + 1/2} + \frac{\sqrt 2 - x}{(x-\sqrt2/2)^2 + 1/2}\right)\end{align*} Next $$\frac{\sqrt 2 \pm x}{(x\pm\sqrt2/2)^2 + 1/2} = \frac{\sqrt 2/2 \pm x}{(x\pm\sqrt2/2)^2 + 1/2} + \frac{\sqrt 2/2}{(x\pm\sqrt2/2)^2 + 1/2}.$$ Now the first term can be integrated by substitution ($u$-sub for the denominator) and the latter by trig-substitution. Indeed, \begin{align*} \int^{S}_0 \frac{\sqrt 2/2 \pm x}{(x\pm\sqrt2/2)^2 + 1/2}dx &= \pm \frac 1 2\log((x\pm\sqrt2/2)^2 + 1/2) \bigg|^{S}_0 \\&= \pm \frac 1 2\log(x^2 \pm x\sqrt2 + 1) \bigg|^{S}_0 \\ &= \pm \frac 1 2\log(S^2 \pm S\sqrt2 + 1)\end{align*} and \begin{align*} \int^S_0 \frac{\sqrt 2/2}{(x\pm\sqrt2/2)^2 + 1/2}dx &= \int^S_0 \frac{\sqrt 2}{2(x\pm\sqrt2/2)^2 + 1}dx \\ &= \tan^{-1}(x\sqrt 2 \pm 1)\bigg|^S_0\\ &= \tan^{-1}(S\sqrt 2 \pm 1) \mp \frac{\pi}{4}. \end{align*} So unraveling all of this using logarithm rules, we have \begin{align*} I^* &= \frac{1}{2\sqrt 2}\bigg(\frac 1 2\log(S^2 + S\sqrt 2 + 1) - \frac 1 2\log(S^2 - S\sqrt 2 + 1) \\ &\hspace{4cm} + \tan^{-1}(S\sqrt2 + 1) + \tan^{-1}(S\sqrt 2 - 1) \bigg)\\ &= \frac{1}{2\sqrt2}\left(\log\left(\sqrt{\frac{S^2 + S\sqrt 2 + 1}{S^2 - S\sqrt 2 + 1}} \right) +\tan^{-1}(S\sqrt2 + 1) + \tan^{-1}(S\sqrt 2 - 1) \right)\end{align*} and thus $$\boxed{I = 1 - \frac{1}{2S\sqrt2}\left(\log\left(\sqrt{\frac{S^2 + S\sqrt 2 + 1}{S^2 - S\sqrt 2 + 1}} \right) +\tan^{-1}(S\sqrt2 + 1) + \tan^{-1}(S\sqrt 2 - 1)\right)}$$ where $\boxed{S = s^{-1/4}(R_x + R_c).}$

And now I leave it to you to check this answer; it's highly possible I made algebra mistakes along the way.

EDIT: pointed out by Claude Leibovici in the comments, using the arctangent addition identity $\tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left( \frac{a+b}{1-ab}\right)$, we can further simplify $$\boxed{I = 1 - \frac{1}{2S\sqrt2}\left(\log\left(\sqrt{\frac{S^2 + S\sqrt 2 + 1}{S^2 - S\sqrt 2 + 1}} \right) +\tan^{-1}\left(\frac{S \sqrt 2}{1-S^2}\right) \right)}$$

Answer 2

After User8128's answer, assuming that you enjoy complex numbers, let $a,b,c,d$ to be the roots of $x^4+1=0$ and write $x^4+1=(x-a)(x-b)(x-c)(x-d)$.

Using partial fraction decomposition, we then have $$\frac 1{x^4+1}=\frac{1}{(a-b) (a-c) (a-d) (x-a)}+\frac{1}{(b-a) (b-c) (b-d) (x-b)}+\frac{1}{(c-a) (c-b) (c-d) (x-c)}+\frac{1}{(d-a) (d-b) (d-c) (x-d)}$$ that is to say $$\frac 1{x^4+1}=\sum_{i=1}^4 \frac {k_i}{x-t_i}\implies I=\int \frac {dx}{x^4+1}=\sum_{i=1}^4 k_i\log(x-t_i)$$

Replacing and recombining, you will end with two logarithms and two arctangents which can be recombined into one logarithm and one arctangent.

$$I=\frac 1{4\sqrt 2}\log \left(\frac{x^2+\sqrt{2} x+1}{x^2-\sqrt{2} x+1}\right)+\frac 1{2\sqrt 2}\tan ^{-1}\left(\frac{x\sqrt{2} }{1-x^2}\right)$$ as already given by User8128.