gerbil Posted April 9, 2012 Share Posted April 9, 2012 Hello all, I'm trying to solve analytically a differential equation and I encountered the following integral, Integrate [1/((Sqrt(x))*(atan(x)), dx] The change of variable u = sqrt(x) brings this integral to, 2 * Integrate [1 / atan(u^2), du] Now what ... ? Any help will be highly appreciated. P.S. I attached a PDF file with the problematic integral. Question.pdf Link to comment Share on other sites More sharing options...
Xittenn Posted April 9, 2012 Share Posted April 9, 2012 [math] \int x^n \; arccot \; \frac{x}{a} \; dx = \frac{x^{n+1}}{n+1} \; arccot \; \frac{x}{a} + \frac{a}{n + 1} \; \int \frac{x^n+1}{a^2 + x^2} \; dx [/math] "Table of Integrals, Series, and Products 7th Ed." - Gradshteyn and Ryzhik Link to comment Share on other sites More sharing options...
D H Posted April 9, 2012 Share Posted April 9, 2012 (edited) Hello all, I'm trying to solve analytically a differential equation and I encountered the following integral, Integrate [1/((Sqrt(x))*(atan(x)), dx] The change of variable u = sqrt(x) brings this integral to, 2 * Integrate [1 / atan(u^2), du] Now what ... ? Your parentheses don't match in 1/((Sqrt(x))*(atan(x)), but given your u-substitution, I gather you are asking about [math]\int \frac 1{\sqrt x\,\arctan x}\,dx[/math] Your integrand is a transcendental function. There is no analytic solution in the elementary functions to this integral. Edited April 9, 2012 by D H Link to comment Share on other sites More sharing options...
gerbil Posted April 10, 2012 Author Share Posted April 10, 2012 Your parentheses don't match in 1/((Sqrt(x))*(atan(x)), but given your u-substitution, I gather you are asking about [math]\int \frac 1{\sqrt x\,\arctan x}\,dx[/math] Your integrand is a transcendental function. There is no analytic solution in the elementary functions to this integral. Yes. That's what I was afraid of. Well, I'll just post the original ODE and maybe you'll have an idea how to solve this one, before I give up. $\dot{x} = - \alpha {x}^{1/2} \arctan{(kx)}$ for $x > 0$ and $t \in [0, T]$ , where $T < \infty$ and $k > 0$. Best, Miki Link to comment Share on other sites More sharing options...
Xittenn Posted April 11, 2012 Share Posted April 11, 2012 (edited) Yes. That's what I was afraid of. Well, I'll just post the original ODE and maybe you'll have an idea how to solve this one, before I give up. [math]\dot{x} = - \alpha {x}^{1/2} \arctan{(kx)}[/math] for [math]x > 0[/math] and [math]t \in [0, T][/math] , where [math]T < \infty[/math] and [math]k > 0[/math]. Best, Miki Just for my reading purposes! On a scale of 1 - 10 in terms of analytically solving first order nonlinear ordinary differential equations how hard is this? Edited April 11, 2012 by Xittenn Link to comment Share on other sites More sharing options...
gerbil Posted April 11, 2012 Author Share Posted April 11, 2012 Just for my reading purposes! On a scale of 1 - 10 in terms of analytically solving first order nonlinear ordinary differential equations how hard is this? I dont think that the word "hard" is relevant here. The problem is that I'm not sure that my equation can be solved analytically (and I'm not sure about the opposite either). Maybe there exists a tricky substitution that solves this equation and maybe there isn't. It is a bummer ... But there are some cases in which nonlinear (first-order) equations can be solved, for example, the Bernoullis' equations. Link to comment Share on other sites More sharing options...
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