Help with Solving Indefinite Integral

[laura1231]

Hi, I tried to solve this integral
$$\int\sqrt{1-\frac{1}{x^3}}dx$$
but i can't solve it...
can someone help me?

 

[MarkFL]

According to wolframalpha, the anti-derivative involves elliptic integrals of the first and second kind. Make certain you have copied the problem correctly, and perhaps also provide some context from which this integral derives. There are folks here who are quite good at advanced integrals, and this information may provide enough context to be useful to them. :D

 

[laura1231]

thanks for your answer, this integral is a consequence of an attempt to solve this differential equation $y''+y-\dfrac{y}{1+x^3}=0$...

 

[chisigma]

thanks for your answer, this integral is a consequence of an attempt to solve this differential equation $y''+y-\dfrac{y}{1+x^3}=0$...

The ODE You have to solve is...

$\displaystyle y^{\ ''} + y\ \frac{x^{3}}{1 + x^{3}}= 0\ (1)$

... isn't it?...

Kind regards

$\chi$ $\sigma$

 

[laura1231]

yes $\chi\sigma$, it seems simple...apparently

 

[chisigma]

yes $\chi\sigma$, it seems simple...apparently

May be it seems... but in fact it is hard!... the second order ODE...

$\displaystyle y^{\ ''} + y\ \frac{x^{3}}{1+x^{3}}\ (1)$

... is linear homogeneous and that means that the solution is of the form...

$\displaystyle y(x) = c_{1}\ u(x) + c_{2}\ v(x)\ (2)$

... where u(*) and v(*) are independent solutions of (1). An asyntotic solution for x>>1 of course is...

$\displaystyle y(x) \sim c_{1}\ \cos x + c_{2}\ \sin x\ (3)$

... but what what does it happen for more small values of x?... one attempt to approximate the solution is to suppose that y(x) is analytic in x=0, so that is...

$\displaystyle y(x)= a_{0} + a_{1}\ x + a_{2}\ x^{2} + ...\ (4)$

For semplicity we suppose also that is $y(0)= 1 \implies a_{0}=1$ and $y^{\ '}(0)=0 \implies a_{1}=0$ so that the problem is to compute the succesive derivatives. From (1) we derive directly...

$\displaystyle y^{\ ''} = - y\ \frac{x^{3}}{1+x^{3}} = 0\ \text{in}\ x=0\ (5)$

Proceeding from (5) we have...

$\displaystyle y^{(3)} = - 3\ y\ \frac{x^{2}}{(1+x^{3})^{2}} - y^{\ '}\ \frac{x^{3}}{1+x^{3}}= 0\ \text{in}\ x=0\ (6) $

$\displaystyle y^{(4)} = y\ \frac{12\ x^{2} - 6\ x}{(1+x^{3})^{3}} - y^{\ '}\ \frac{6\ x^{2}}{(1+x^{3})^{2}} - y^{\ ''}\ \frac{x^{3}}{1+x^{3}} = 0\ \text{in}\ x=0\ (7) $

$\displaystyle y^{(5)} = - 6\ y\ \frac{14\ x^4 - 8\ x^{3} - 4\ x + 1}{(1 + x^{3})^{4}} + ... = -6\ \text{in}\ x=0\ (7)$

If we stop now the approximate solution near x=0 is...

$\displaystyle y(x) \sim 1 - \frac{x^{5}}{20}\ (8)$

If more precision is requested we can go ahead but the task is very inconfortable... at least for me (Worried)... A numerical solution is clearly preferable...

Kind regards

$\chi$ $\sigma$

 

[laura1231]

Thanks $\chi\ \sigma$, in effect this equation is very hard to solve... I think that is impossible to find an analytic solution.. The numerical solution seems to be the only way at moment...