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

[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

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, 201213 yr by D H

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

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, 201213 yr by Xittenn

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.