Why can't a value for this definite integral be found?

Question

I was trying to find out if $\int _0^{\infty }\:\frac{1}{\left(1+x^3\right)^{\frac{1}{2}}}dx$ converges or diverges. I split it into a sum, that is

$\int _0^{1 }\:\frac{1}{\left(1+x^3\right)^{\frac{1}{2}}}dx$ + $\int _1^{y }\:\frac{1}{\left(1+x^3\right)^{\frac{1}{2}}}dx$

where $y>1$ . I found that the second integral of the sum converges as y tends to infinity, but I can not evaluate the first integral. When I plot the graph, I can see clearly that the area under the curve of $f(x)=\frac{1}{\left(1+x^3\right)^{\frac{1}{2}}}$ with x in $[0,1]$ is finite. What is wrong with that integral?

You don't need to evaluate it to determine whether or not it converges. It's easy enough to just say that $(1+x^2)^{-1/2}$ is continuous on $[0,1]$ and therefore the first integral exists. — Antonio Vargas, Nov 12 '14 at 23:59
I know I don't need to evaluate it, I just wanted to. But thanks Antonio. Elliptic integral? Ok — 2good4this, Nov 13 '14 at 00:01
According to mathworld, the integral has the value $2.804364...$ — Peter, Nov 13 '14 at 00:01
If you want the solution only, it is $\frac{2\Gamma(1/3)\Gamma(7/6)}{\sqrt{\pi}}$ — dustin, Nov 13 '14 at 00:01
The first integral is an elliptic integral of the first kind. That's why you have a hard time evaluating it. — UserX, Nov 13 '14 at 00:03

score 2 · Answer 1 · answered Nov 13 '14 at 03:11

2

Hint: Let $~t=\dfrac1{1+x^3}~$ and then recognize the expression of the beta function in the new integral.

answered Nov 13 '14 at 03:11

Lucian

48,334
2
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score 0 · Answer 2 · answered Nov 13 '14 at 00:06

0

Let $u^2 = 1+x^3 \to 2udu = 3x^2dx = 3(u^2-1)^{2/3}dx \to I = \displaystyle \int_{1}^{\sqrt{2}} \dfrac{2}{3\left(u^2-1\right)^{2/3}}du$. Next you can make a substitution: $u = \sec \theta$.

answered Nov 13 '14 at 00:06

DeepSea

77,651

Why can't a value for this definite integral be found?

2 Answers2