Recent content by Random Variable

Now I'm not so sure if [0, \infty) is considered to be a closed interval or not. But regardless of that, do we also have an issue at x=0 even if f(x) = 0 ?

But even if \sum \ln x \ x^{n} did converge uniformly for x=1 we would still have a problem at x=0, right? And since we are not integrating over a closed interval, we couldn't invoke uniform convergence to justify that \int_{0}^{\infty} \frac{f(x)}{1+e^{x}} \ dx = \int_{0}^{\infty}...

I'm a bit confused. You said that we need to show that \int_0^1\sum_{n=1}^{+\infty}|ln(x)x^n| dx<+\infty. I originally showed that \sum_{n=1}^{+\infty}\int_0^1|ln(x)x^n| dx<+\infty (because that's what it says needs to be shown on the web page to which you linked). But since one implies...

Is something like this what you had in mind? \int_0^1 \frac{|ln(x)|}{1-x}dx = \int_0^\frac{1}{2} \frac{|ln(x)|}{1-x}dx + \int_\frac{1}{2}^{1} \frac{|\ln (x)|}{1-x} \ dx \int_0^\frac{1}{2} \frac{|ln(x)|}{1-x}dx < \int_{0}^{\frac{1}{2}} \frac{1}{(1-x)\sqrt{x}} \ dx < \infty since the integral...

Thanks. I'm familiar with both convergence theorems, although my knowledge of Lebesgue integration is limited. I have one more question. If \sum_{n=0}^{\infty} x^{n} does not converge for x = 1 , why is it necessarily true that \int_{0}^{1} \frac{f(x)}{1-x} \ dx = \int_{0}^{1} f(x)...

So you're saying that I didn't actually have to evaluate it directly. I just needed to show that it converges. Do you have a link for that second criterion? Is it also something from measure theory?

Could you say the following? \int_{0}^{1} |\ln x \ x^{n} | \ dx = - \int_{0}^{1} \ln x \ x^{n} = \frac{1}{(n+1)^{2}} and \sum_{n=0}^{\infty} \frac{1}{(n+1)^{2}} = \zeta(2) = \frac{\pi}{6} < \infty

Say that f(x)= \ln x .

Homework Statement I want to justify that \int_{0}^{1} \frac{f(x)}{1-x} \ dx = \int_{0}^{1} f(x) \sum_{k=0}^{\infty} x^{n} \ dx = \sum_{k=0}^{\infty} \int_{0}^{1} f(x) x^{n} \ dx Homework Equations The Attempt at a Solution I always thought changing the order of summation...

I want to evaluate the integral without using a closed contour and the residue theorem.

Is there a problem with the following evaluation?\displaystyle \int e^{-ix^{2}} \ dx = \frac{1}{\sqrt{i}} \int e^{-u^{2}} \ du = \left( \frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \text{erf}(u) + C = \left(\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) \text{erf} (\sqrt{i}x) + C So...

Can you informally say that if A is a 4x4 matrix whose column space spans all of R^4, a basis for the null space of B (a 2x4 matrix with null space of dimension 2) will also form a basis for space X?

Thanks for answering all my questions. What am I going to do next may seem absurd, but I just wanted to make sure I understand the concept. If there were no restrictions on y, then y could be any vector in R^4. So a basis of y would be {e1,e2,e3,e4}. Then the vector space in question would...

I guess what I'm trying to say is that isn't it possible for different matrices A and B that A*(basis vectors of null(B)) could just be a spanning set and not a basis for x?

What if A were rank deficient? You couldn't automatically say that A*{b1,b2} (where b1 and b2 are the basis vectors of null(B)) is a basis for x? Wouldn't you have to check first?