# Convergence of Improper Integrals: Homework

• Mattofix
Can you think of any expressions that would work for both i) and ii)? Or, can you find a way to manipulate your current expressions so that the inequalities are reversed? Remember, you can multiply or divide both sides of an inequality by a positive number without changing its direction.
Mattofix

## Homework Statement

do the following integrals converge?

i) $$\int_0^{1}\frac{dx}{x^{3/2}e^{x}}$$

ii) $$\int_0^{1}\frac{x}{\sqrt{1-x^{2}}}dx$$

## The Attempt at a Solution

looking at them i can guess that they both diverge - proving this is the hard part - this is what i have got but it doesn't prove anything...

i) $$\frac{1}{x^{3/2}}$$$$\geq1$$

$$\frac{1}{x^{3/2}e^{x}}$$$$\geq{e^{-x}}$$

$$e^{-x}$$ converges
$$\frac{1}{e^{x}}$$$$\leq1$$

$$\frac{1}{x^{3/2}e^{x}}$$$$\leq{x^{-3/2}$$

$$x^{-3/2}$$ divereges
ii) $$x\leq1$$

$$\frac{x}{\sqrt{1-x^{2}}}$$$$\leq{\frac{1}{\sqrt{1-x^{2}}}$$

$${\frac{1}{\sqrt{1-x^{2}}}$$ diverges (cannot prove it though)
$${\frac{1}{\sqrt{1-x^{2}}}\geq1$$

$${\frac{x}{\sqrt{1-x^{2}}}$$$$\geq x$$

x converges

any help would be much appreciated.

Last edited:
Mattofix said:
i) $$\frac{1}{x^{3/2}}$$$$\geq1$$

$$\frac{1}{x^{3/2}e^{x}}$$$$\geq{e^{-x}}$$

$$e^{-x}$$ converges

I take your last statement above to mean that when you integrate both sides of the inequality with respect to x from 0 to 1, the right-hand side converges. That is true. Now what does that tell you about the left-hand side?

$$\frac{1}{e^{x}}$$$$\leq1$$

$$\frac{1}{x^{3/2}e^{x}}$$$$\leq{x^{-3/2}$$

$$x^{-3/2}$$ divereges

Again, what does that tell you about the left-hand side? Ditto for ii).

e(ho0n3 said:
I take your last statement above to mean that when you integrate both sides of the inequality with respect to x from 0 to 1, the right-hand side converges. That is true. Now what does that tell you about the left-hand side?

maybe I'm missing something but not much because the left hand side is greater than the right, the right converges, this does not mean the same for the left - it might diverge.

The same for all the others - the comparisions don't tell us much.

If all the inequalities were reversed then i wouldn’t have a problem.

i think...?

Last edited:
Yes. Exactly. You need to find an expression so that the inequalities are reversed.

## Related to Convergence of Improper Integrals: Homework

H2 - What is the definition of an improper integral?

An improper integral is an integral where one or both of the bounds are infinite or the integrand has an infinite discontinuity within the bounds. It is also called a divergent integral as it may not have a finite value.

H2 - How is the convergence of an improper integral determined?

The convergence of an improper integral is determined by evaluating the limit of the integral as the bounds approach infinity or the point of discontinuity. If the limit exists and is finite, then the integral is said to converge. If the limit does not exist or is infinite, then the integral is said to diverge.

H2 - What are the different types of convergence for improper integrals?

The two main types of convergence for improper integrals are absolute convergence and conditional convergence. Absolute convergence means that the integral converges regardless of the order in which the limits are taken. Conditional convergence means that the integral only converges if the limits are taken in a specific order.

H2 - How do we handle integrals with infinite bounds?

Integrals with infinite bounds can be handled by taking the limit of the integral as the bounds approach infinity or negative infinity. If the limit exists and is finite, then the integral converges. If the limit does not exist or is infinite, then the integral diverges.

H2 - Can improper integrals be solved using the fundamental theorem of calculus?

Yes, improper integrals can be solved using the fundamental theorem of calculus. However, the integrals must first be evaluated as a limit before applying the theorem. If the limit does not exist, then the theorem cannot be applied and the integral is said to diverge.

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