Improper Integrals from a Calculus BC Class.

Moriae

Salutations,

I have had some troubles with improper integrals. After the lesson with my teacher and reading the math book, I still don't understand precisely what to do. I'm using James Stewart's Calculus: Early Transcendentals I have a question on how to evaluate both continuous and discontinuous improper integrals. Here are two examples from the book I'm using.

Continuous: Evaluate the integral x*e^x with respect to x from negative infinity to zero.

Discontinuous: Evaluate the integral 1/(x-1) with respect to x from zero to three if possible.

Any advice, guidance, or help on this would be much appreciated. I think I might be missing one key thing, but I am not certain.

Thank you in advance,
Brandon[/i]

**Posted:** Sun Feb 06, 2005 3:43 pm

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Bictor717

I haven't done improper integrals yet, but I'll give it a shot based on what I know from reading ahead. You should probably wait until someone more adept comes along. I also used integration by parts which we haven't covered in class either.
1.
$\int^0_{-\infty}xe^x\;dx=\lim_{n\to -\infty}\int^0_n xe^x\;dx$
$=\lim_{n\to -\infty}\displaystyle\left(\displaystyle\left.xe^x\right]^0_n-\int^0_n e^x\;dx\right)$
$=\lim_{n\to -\infty}\displaystyle\left(0-ne^n\right)-\lim_{n\to -\infty}\displaystyle\left(e^0-e^n\right)$
$=\lim_{n\to -\infty}\left(e^n-ne^n-1\right)$
$=-1$
2.
$\frac{1}{x-1}$ is symmetric about (1,0), so $\int^0_2 \frac{1}{x-1}\;dx=0$ .
$\int^0_3 \frac{1}{x-1}\;dx=\int^2_3 \frac{1}{x-1}\;dx=\ln 2$ .

**Posted:** Sun Feb 06, 2005 6:17 pm

Moriae

Thank you very much,

The first one you were very correct on. The second one was not correct, but I don't understand where you really went wrong.

Brandon

**Posted:** Sun Feb 06, 2005 8:00 pm

_________________
The Australian Minister of Labour once said in 1973 "We hope that one day everyone will be paid above the average wage rate."

TripleM

**Posted:** Sun Feb 06, 2005 10:38 pm

fedja

A lot depends on the definition you want to use here. For the case of finite singularity s inside the interval [a,b] of integration you may want to undestand the improper integral as the limit of the sum of integrals from a to s-d1 and from
s+d2 to b when d1,d2>0 and tend to 0 independently. If so, then the integral, indeed, does not exist. Another possibility is to consider the so-called principal value when you take d1=d2=d>0 and let d tend to 0. In this case the limit exists and is ln2.

**Posted:** Sun Feb 06, 2005 10:56 pm

Kent Merryfield

First: you can post questions like this over in the "Caclulus calculations and tutorials" forum under "College playground."

Fedja mentioned the principal value. I'd mention two caveats:

1. It's not a default of the notation. If you intend to be talking about the principal value, you'd better say that somewhere. If you just say $\int_0^3\frac{dx}{x-1}$ without further comment, then you do not mean the principal value.

2. I just looked up "principal value" in the index of Stewart. (Not the current edition, but that probably doesn't matter.) It's not there. There's no doubt that Stewart did not intend this problem to involve the concept.

Here's a framework for looking at this in the case of any individual integral: for how many distinct reasons is this integral improper? $\int_{-\infty}^0xe^{-x}dx$ is improper for one reason: the $-\infty.$ But $\int_0^3\frac{dx}{x-1}$ is improper for two reasons: the left side of 1 and the right side of 1.

The rule is that you must separate the integral into pieces each of which is improper for only one reason. In order to claim convergence, you must separately have convergence on each piece. In the case of $\int_0^3\frac{dx}{x-1}$ one piece is $-\infty$ and the other is $+\infty$ , so the integral is divergent.

Another example: $\int_{-\infty}^{\infty}\frac x{x^2+1}\,dx$ is improper for two reasons. Since $\int_0^{\infty}\frac x{x^2+1}\,dx=\infty$ (Note comparison to the integral of $\frac1x$ or the harmonic series), the whole integral diverges despite its symmetry.

And any student who tries to tell me that $\int_0^3\frac{dx}{(x-1)^2}=-\frac32$ will be asked to explain how they think a positive function could give a negative integral.

**Posted:** Tue Feb 08, 2005 8:46 pm

joml88

**Posted:** Tue Feb 08, 2005 8:50 pm

fedja

**Posted:** Tue Feb 08, 2005 11:16 pm