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?
$\endgroup$ 04 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$Convergent. Compare with $\int_0^1 \dfrac{1}{x^{1/2}}\; dx$.
$\endgroup$ $\begingroup$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.
$\endgroup$ 3 $\begingroup$$$\frac{1}{\sqrt{x+x^5}}=\frac{1}{\sqrt{x}\sqrt{1+x^4}}$$
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:
$$\int_0^1\frac{\mathrm{d}x}{\sqrt{x+x^5}} = 2 \times\, _2 F_1\left(\frac{1}{8},\frac{1}{2};\frac{9}{8};-1\right) \approx 1.91773$$
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