# Help with improper integrals.

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• October 10th 2012, 06:46 AM
SecondxChance1
Help with improper integrals.
So I'm having some trouble with quite a few problems in this section. I'm fairly certain my problem is just setting them up. I think I'm forgetting some of the concepts learned in Calc 1, and that's what's giving me trouble. That said, if someone could point out what exactly I'm doing wrong, that would be awesome.

One of the problems I'm having trouble wants me to determine if the integral is convergent or divergent.

The integral is: The integral from negative infinity to -1 of 1 over the square root of (1-w) dw

My problem is the radical in the denominator. I'm not sure how to get rid of it.
• October 10th 2012, 07:30 AM
emakarov
Re: Help with improper integrals.
Using the substitution x = 1 - w, we get $\int_{-\infty}^{-1}\frac{1}{\sqrt{1-w}}\,dw=\int_2^\infty\frac{1}{\sqrt{x}}\,dx$. Find the second integral when the upper bound is a finite number y and see what it tends to when $y\to\infty$.
• October 10th 2012, 11:43 AM
SecondxChance1
Re: Help with improper integrals.
Thanks! I managed to figure it out now.
However, I'm stumped on another question now. Same situation as before, where I need to figure out whether it's converging or diverging.

This time it's the integral from negative infinity to positive infinity of x e^-x^2 dx

I tried letting u = -x^2, du = -2dx

Which gave me -1/2 int e^u du and that integrates into just -1/2e^u

I split the limit up at 0, so I had the limit from negative infinity to zero and from zero to positive infinity. I tried working it out from here, but never managed to get the right answer. What I would like to know is this first part is right, and if not, how should I set up this problem?
• October 10th 2012, 11:52 AM
TheEmptySet
Re: Help with improper integrals.
Quote:

Originally Posted by SecondxChance1
Thanks! I managed to figure it out now.
However, I'm stumped on another question now. Same situation as before, where I need to figure out whether it's converging or diverging.

This time it's the integral from negative infinity to positive infinity of x e^-x^2 dx

I tried letting u = -x^2, du = -2dx

Which gave me -1/2 int e^u du and that integrates into just -1/2e^u

I split the limit up at 0, so I had the limit from negative infinity to zero and from zero to positive infinity. I tried working it out from here, but never managed to get the right answer. What I would like to know is this first part is right, and if not, how should I set up this problem?

You are the right path.

$\int_{-\infty}^{\infty}xe^{-x^2}dx$

So you broke this up into two integrals

$\int_{-\infty}^{0}xe^{-x^2}dx+\int_{0}^{\infty}xe^{-x^2}dx$

Now let $u=-x^2 \implies du=-2xdx \iff \frac{-du}{2}=xdx$

$-\int_{-\infty}^{0}\frac{e^{u}}{2}du-\int_{0}^{-\infty}\frac{e^{u}}{2}du$

Now if you use the property that

$\int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx$

You will get the correct answer 0.

Note: the function

$f(x)=xe^{-x^2}$ is odd

$f(-x)=-xe^{-(-x)^2}=-xe^{-x^2}=-f(x)$

The integral of an odd function over a symmetric interval is always zero.
• October 11th 2012, 07:28 AM
SecondxChance1
Re: Help with improper integrals.
Okay, thanks.

I've got 2 more I don't know how to start though. Same type of problem as before, just need to figure out whether it's divergent or convergent.

integral from -2 to 14 of 1/4th root of (x+2) dx

and the integral from -2 to 3 of 1/x^4 dx

I've tried a few things, but if I could get a little clue as to how I'm supposed to set them up, I should be able to solve them.
• October 11th 2012, 08:10 AM
emakarov
Re: Help with improper integrals.
Quote:

Originally Posted by SecondxChance1
integral from -2 to 14 of 1/4th root of (x+2) dx

and the integral from -2 to 3 of 1/x^4 dx

The method is the same as in post #2. The general fact is that $\int_0^1 x^a\,dx$ converges if a < 1 and $\int_1^\infty x^a\,dx$ converges if a > 1.
• October 12th 2012, 05:14 AM
emakarov
Re: Help with improper integrals.
There is a typo in post #6. It should say
Quote:

Originally Posted by emakarov
The general fact is that $\int_0^1 x^a\,dx$ converges iff a > -1 and $\int_1^\infty x^a\,dx$ converges iff a < -1.