3
$\begingroup$

Find this integral $$\int\dfrac{\sqrt{x-1}\arctan{(x\sqrt{x-1})}}{x}dx$$

My try: let $$\arctan{(x\sqrt{x-1})}=t$$ and that's very ugly,Thank you

$\endgroup$
  • $\begingroup$ Mathematica can't find a primitive at all. $\endgroup$ – Lucian Dec 2 '13 at 10:09
  • $\begingroup$ I would really like to learn how to compute this antiderivative ! $\endgroup$ – Claude Leibovici Dec 2 '13 at 10:13
1
$\begingroup$

This problem is not really very hard yet it is messy since one needs to deal with roots of third order polynomials. Yet all what is involved in here is one substitution, then integration by parts and then decomposition into partial fractions. We have: \begin{equation} \int \frac{\sqrt{x-1}}{x} \arctan\left[ x \sqrt{x-1}\right] dx = \int\frac{2 u^2}{1+u^2} \arctan\left[ u(1+u^2)\right] du \end{equation} where we substituted for $u=\sqrt{x-1}$. Now we integrate by parts: \begin{eqnarray} \int\frac{2 u^2}{1+u^2} \arctan\left[ u(1+u^2)\right] du = 2(u-\arctan(u)) \cdot \arctan\left[ u(1+u^2) \right] - 2 \int\frac{u-\arctan(u)}{1+(1+u^2)^2 u^2} (1+3 u^2) du \end{eqnarray} where we used the fact that $\int (2 u^2)/(1+u^2) du = 2(u-\arctan(u))$.

Now there are two integrals to be computed .

Let us start from the easier one: \begin{eqnarray} &&\int \frac{u(1+3 u^2)}{1+(1+u^2)^2 u^2} du =\\ && u \arctan\left[ u(1+u^2)\right] - \int \arctan\left[ u(1+u^2) \right]du=\\ && u \arctan\left[ u(1+u^2)\right] -\frac{\pi}{2} u- \frac{1}{2 \imath} \sum\limits_{\xi=1}^3 \log(u-u^{(+)}_\xi) (u-u^{(+)}_\xi)+ \frac{1}{2 \imath} \sum\limits_{\xi=1}^3 \log(u-u^{(-)}_\xi) (u-u^{(-)}_\xi) \end{eqnarray} where in the last line we used the identity $\arctan(x)= 1/(2 \imath) \log[(1+\imath x)/(1-\imath x)]$.

Here $\left\{ u^{(+)}_\xi\right\}_{\xi=1}^3$ are roots of the equation $u^3+u-\imath=0$ and $\left\{ u^{(-)}_\xi\right\}_{\xi=1}^3$ are roots of the equation $u^3+u+\imath=0$.

Now the second integral: \begin{eqnarray} &&\int\frac{(1+3 u^2) \arctan[u]}{1+(1+u^2)^2 u^2} du=\\ &&\arctan[u(1+u^2)] \arctan[u] - \frac{\pi}{2} \arctan[u] + \frac{1}{4} \sum\limits_{\xi=1}^3 \sum\limits_{\eta \in\{-1,1\}} \sum\limits_{\eta_1 \in\{-1,1\}} F^{(u^{(\eta)}_\xi,-\eta_1 \imath)}(u)\\ \end{eqnarray} where \begin{equation} F^{(a,b)}(u):=\log(a+u) \log[\frac{x+b}{b-a}] + Li_2\left[ \frac{x+a}{a-b}\right] \end{equation}

Bringing everything together we have: \begin{eqnarray} &&\int\frac{\sqrt{x-1}}{x} \arctan\left[ x \sqrt{x-1}\right] dx=\pi \left( \sqrt{x-1} - \arctan[\sqrt{x-1}]\right)+\\ && -\imath \sum\limits_{\xi=1}^3 \left(\log[\sqrt{x-1}-u^{(+)}_\xi](\sqrt{x-1}-u^{(+)}_\xi)- \log[\sqrt{x-1}-u^{(-)}_\xi](\sqrt{x-1}-u^{(-)}_\xi)\right)+\\ &&\frac{1}{2} \sum\limits_{\xi=1}^3 \sum\limits_{\eta \in\{-1,1\}} \sum\limits_{\eta_1 \in\{-1,1\}} F^{(u^{(\eta)}_\xi,-\eta_1 \imath)}(\sqrt{x-1}) \end{eqnarray}

In[1296]:= x =.;
eX = D[Pi Sqrt[x - 1] - Pi ArcTan[Sqrt[x - 1]] - 
     I Sum[Log[
          Sqrt[x - 1] - Root[#^3 + # - I &, xi]] (Sqrt[x - 1] - 
           Root[#^3 + # - I &, xi]) - 
        Log[Sqrt[x - 1] - Root[#^3 + # + I &, xi]] (Sqrt[x - 1] - 
           Root[#^3 + # + I &, xi]), {xi, 1, 3}] + 
     1/2 Sum[eta eta1 (F[-Root[#^3 + # - eta I &, xi], -eta1 I, Sqrt[
          x - 1]]), {xi, 1, 3}, {eta, -1, 1, 2}, {eta1, -1, 1, 2}], 
    x] - Sqrt[x - 1]/x ArcTan[x Sqrt[x - 1]];
x = RandomReal[{0, 10}, WorkingPrecision -> 50];
N[eX, 50]

Out[1299]= 0.*10^-50 + 0.*10^-50 I
$\endgroup$
-5
$\begingroup$

$\because$ according to http://integrals.wolfram.com/index.jsp?expr=%28x-1%29%5E%281%2F2%29%2Fx&random=false, $\int\dfrac{\sqrt{x-1}}{x}dx=2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1}+C$

$\therefore\int\dfrac{\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})}{x}dx$

$=\int\tan^{-1}(x\sqrt{x-1})~d(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})$

$=(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})\tan^{-1}(x\sqrt{x-1})-\int(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})~d(\tan^{-1}(x\sqrt{x-1}))$

$\because$ according to http://www.wolframalpha.com/input/?i=d%2Fdx%28arctan%28x%28x-1%29%5E%281%2F2%29%29%29, $\dfrac{d}{dx}(\tan^{-1}(x\sqrt{x-1}))=\dfrac{3x-2}{2\sqrt{x-1}(x^3-x^2+1)}$

$\therefore(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})\tan^{-1}(x\sqrt{x-1})-\int(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})~d(\tan^{-1}(x\sqrt{x-1}))$

$=2\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})-2\tan^{-1}\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})-\int(2\sqrt{x-1}-2\tan^{-1}\sqrt{x-1})\dfrac{3x-2}{2\sqrt{x-1}(x^3-x^2+1)}dx$

$=2\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})-2\tan^{-1}\sqrt{x-1}\tan^{-1}(x\sqrt{x-1})-\int\dfrac{3x-2}{x^3-x^2+1}dx+\int\dfrac{(3x-2)\tan^{-1}\sqrt{x-1}}{\sqrt{x-1}(x^3-x^2+1)}dx$

$\int\dfrac{3x-2}{x^3-x^2+1}dx$ is just an integral of rational functions and should have close form.

For $\int\dfrac{(3x-2)\tan^{-1}\sqrt{x-1}}{\sqrt{x-1}(x^3-x^2+1)}dx$ ,

Let $u=\sqrt{x-1}$ ,

Then $x=u^2+1$

$dx=2u~du$

$\therefore\int\dfrac{(3x-2)\tan^{-1}\sqrt{x-1}}{\sqrt{x-1}(x^3-x^2+1)}dx$

$=\int\dfrac{2(3u^2+1)\tan^{-1}u}{(u^2+1)^3-(u^2+1)^2+1}du$

Then you can separate it to the terms of $\int\dfrac{\tan^{-1}u}{au+b}du$ or $\int\dfrac{\tan^{-1}u}{pu^2+qu+r}du$ by partial fraction. According to http://pi.physik.uni-bonn.de/~dieckman/IntegralsIndefinite/IndefInt.html, they relate to the polylogarithm function.

$\endgroup$
  • 4
    $\begingroup$ this is unreadable $\endgroup$ – Norbert Dec 5 '13 at 17:06
  • $\begingroup$ Even that integral of rational function is not elementary, so (-1) $\endgroup$ – Norbert Dec 5 '13 at 17:15
  • $\begingroup$ @Norbert: I don't agree with you. For partial fraction using in integration, it is not a must to factorize in $\mathbb{Q}$ , factorize in $\mathbb{R}$ or even in $\mathbb{C}$ is also allowed. The downvotes should be unreasonable. $\endgroup$ – Harry Peter Dec 9 '13 at 14:15
  • $\begingroup$ Then please provide us with explicit formula for the antiderivative $\endgroup$ – Norbert Dec 9 '13 at 14:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.