Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

1 Answer 1

up vote 0 down vote accepted

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.