Does this integral converge or diverge? $\int_{16}^{500} \frac{1}{x^{0.25} - 2} dx,$

Question

I have the $$\int_{16}^{500} \frac{1}{x^{0.25} - 2} dx,$$ and am trying to find whether it converges or diverges.

I have sketched the graph and noticed that their is an asymptote at $x=16$ (hence why the integral is improper for these boundaries).

I am now trying to evaluate the limits to see if it converges or diverges but I am unsure how to approach this. Does anyone have any suggestions or hints? Thanks!

Answer 1

Do the change of variable $t = x^{\frac{1}{4}}-2$, i.e. $x = (t+2)^4$. Then $$ \int_{16}^{500} \frac{dx}{x^{\frac{1}{4}}-2} = \int_{0}^{500^{\frac{1}{4}}-2} \frac{4(t+2)^3}{t}dt $$ and around $0$ the integrand is equivalent to $\frac{32}{t}$ -- is it integrable?

Answer 2

Let $x=t^4$

Then, $dx=4t^3dt$

Now, we have:

$I=\int{ \frac{1}{x^{0.25}-2}dx}=\int{\frac{4t^3}{t-2}dt}$

We use long division to get,

$I=\int{4t^2+8t+\frac{32}{t-2}+16dt}=\frac{4}{3}t^3+4t^2+32\ln(t-2)+16t$

We can ignore the constant of integration since we're going to end up evaluating a definite integral anyway.

Now, as $x\to16$, $t\to2$. As you can see, this is easily seen to tend to $-\infty$ and thus the improper integral is divergent.

Answer 3

Let $b > 16.$ We claim that $\int_{b}^{100}\frac{1}{x^{1/4} - 2} dx$ does not converge as $b \to 16+.$

But, since $$\int \frac{1}{x^{1/4} - 2} dx \simeq \frac{4x^{3/4}}{3} + 4x^{1/2} + 16x^{1/4} + 32 \log |2 - x^{1/4}|,$$ we have $$\frac{-4b^{3/4}}{3} - 4b^{1/2} - 16b^{1/4} + 32 \log |2 - b^{1/4}| \to \frac{-32}{3} - 16 - 32 + 32\log 0$$ as $b \to 16+.$ Thus no convergence follows.