Question

Verify if the integral below converts :

Answer 1

\[\int\limits_{-\infty}^{\infty} (1/(x^3+1))*dx \]
I had calculate that's converts to 0, But I'm not sure. Some one can give a shot ? =)

Answer 2

1-x3+x6-x9+x12-x15 .....
------------------
1+x3 ) 1
(1+x3)
-------
-x3
(-x3-x6)
---------
x6 .....

\[\int_{-inf}^{inf} 1-x3+x6-x9+x12-x15 .....\ dx\]


\[\lim_{x\to inf} \frac{(-1)^nx^{3x+1}}{3x+1}-\lim_{x\to -inf} \frac{(-1)^nx^{3x+1}}{3x+1}\]

Answer 3

im sure i messed up some notation somewhere tho

Answer 4

o.0 Well, I used the comparation test with \[(1/(x^3+1)) \le (1/x)\]

Than i cheked that this integral convert to 0. Can I do this way ?

Answer 5

\[\int( 1-x^3+x^6-x^9+x^{12}-x^{15} ...)\ dx=\]\[\hspace{5em }x-\frac{1}{4}x^4+\frac{1}{7}x^7-\frac{1}{10}x^{10}+\frac{1}{13}x^{13}-\frac{1}{16}x^{16} ...\]

well, ida used the comparison of 1/x^3 if you go that route

Answer 6

comparison test is for discrete values; and integration itself is much larger by nature

Answer 7

It doesn't converge. It diverges at \(x=-1\)

Answer 8

\[\int\limits_{-1}^{-0.99} \frac{1}{x^3+1}dx =\int\limits_{-1}^{-0.99} \frac{1}{x+1}\frac{1}{x^2-x+1}dx>\frac{1}{8}\int\limits_{-1}^{-0.99} \frac{1}{x+1}dx=\frac{1}{8}\int\limits_{0}^{0.01} \frac{1}{x}dx\]
and the last integral diverges