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Check convergence of these improper integrals:

a) $\displaystyle \int_{0}^{+\infty}x^{17}e^{-\sqrt{x}}\mbox{d}x$

b) $\displaystyle \int_{0}^{1}\frac{\sin x}{x^{3/2}} \mbox{d}x$

c) $\displaystyle \int_{0}^{1}\frac{1}{\sqrt[3]{1-x^3}} \mbox{d}x$

I know comparision test and Dirichlet test. But still have troubles using them. Any hints for these examples? I want to practise.

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1 Answer 1

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For a):

You could first make a substitution $u=\sqrt x$. The integral then becomes $\int_0^\infty 2u^{35} e^{-u}\,du$, which is easier to deal with. Here, towards making a comparison, show that for $u$ sufficiently large, $u^{35}<e^{ u/2}$.

For b):

Use the inequality $0\le \sin x\le x$ for $x\ge0$.

For c):

You could first make a substitution: $u=1-x^3$. This gives the integral $$ {1\over3}\int_0^1 { du\over u^{1/3} (1-u)^{2/3} } = {1\over3}\int_0^{1/2} { du\over u^{1/3} (1-u)^{2/3} } +{1\over3}\int_{1/2}^1 { du\over u^{1/3} (1-u)^{2/3} }.$$ You should be able to deal with the latter two integrals (by essentially neglecting one of the terms in the denominator for each).

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