Convergence of an integral $\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
$\int_1^{+\infty} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx=\lim\limits_{t\to\infty}\int_1^{t} \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$
Partial integration can't solve the integral $\int \frac{1}{x\sqrt[3]{x^2+1}}\mathrm dx$.
What substitution (or other methods) would you suggest?
Answer
For $x \in [1,\infty)$, you have $$0 \le \frac{1}{x\sqrt[3]{x^2+1}} \le \frac{1}{x^\frac{5}{3}}$$ and $\int_1^\infty \frac{dx}{x^\frac{5}{3}}$ is convergent.
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