Convergent or Divergent Integral

Question

Convergent or Divergent?

$$\int_0^1 \frac {dx}{(x+x^{5})^{1/2}} $$

I have problem with the fact that if we have integration from 0 to a say and a to infinity. How does this change the way we do the comparison test ?

I have in my textbook: If we have from 0 to a and compare with the function

$$ \frac {1}{(x^{3})} $$ then the exponent of x should be less than 1 for it too be convergent. Which I would say is Divergent. But I am wrong?

Answer 1

Outline: Note that $\sqrt{x+x^4}\ge x^{1/2}$ in our interval. Now recall that $\int_0^1 \frac{dx}{x^{1/2}}$ converges.

Answer 2

Convergent. Compare with $\int_0^1 \dfrac{1}{x^{1/2}}\; dx$.

Answer 3

$$\frac{1}{\sqrt{x+x^5}}=\frac{1}{\sqrt{x}\sqrt{1+x^4}}$$

Near $x\approx0$ you have $1+x^4 \approx 1$. And the integral of $1/\sqrt{x}$ converges near 0. Therefore your integral converges.

Moreover, the result, as computed by Mathematica, is:

$$\int_0^1\frac{\mathrm{d}x}{\sqrt{x+x^5}} = 2 \times\, _2 F_1\left(\frac{1}{8},\frac{1}{2};\frac{9}{8};-1\right) \approx 1.91773$$

Answer 4

Set $x=y^2$, then $dx=2ydy$. So the integral becomes:

$$I=\int_0^1 \frac {1}{(x+x^{5})^{1/2}}dx=\int_0^1 \frac {2y}{(y^2+y^{10})^{1/2}}dy=\int_0^1 \frac {2}{(1+y^5)^{1/2}}dy$$

Thus the integral is convergent.