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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?

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up vote 3 down vote accepted

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.

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

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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.

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+1 I like this solution because you changed an improper integral (it's always a question what that will do) to a regular proper integral which is evidently convergent! – imranfat Aug 20 '14 at 15:23
Thanks!. By the way, $dx$ was not explicitly showed up in the original question! – mike Aug 20 '14 at 15:33
I edited, because it should be there... – imranfat Aug 20 '14 at 15:56


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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Is that an analytic antiderivative? And (I know, I should have faith in Mathematica), do we know that the numerical solution to the $_2F_1$ function is valid over the defined range? – Carl Witthoft Aug 20 '14 at 16:08
@CarlWitthoft The analytic antiderivative returned by Mathematica is somewhat complicated. In the particular integration range from 0 to 1 it simplifies to the result I posted. If someone does it by hand it would be welcome. – becko Aug 20 '14 at 18:10

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