Improper Integral Question/Check

[moo5003]

Hello, I just finished up two problems for my homework and I have a sneaking notion that I have made a mistake somewhere because when I checked the answer numerically by calculator and I get a differing number.

I'm doing improper integrals for my real analysis class and the problem is stated as:

Calculate a) Integral from 0 to 1 of log(x)

Work: Derivative of (xlog(x)-x) = log(x) thus: let 0<d<1

Lim d-> 0 (from the right) Integral from d to 1 of log(x) = Lim d -> 0 xlog(x)-x evaluated from d to 1 = lim d -> 0 of d-1-dlog(d) = 0 - 1 - 0*-Infinity = -1

Integral from 0 to 1 of Log(x) = -1

Though, when i do this on my calculator i get something around -.43.

Similarly I get:

Integral from 2 to Infinity of Log(x)/x = -log^2(2)/2
Integral from 0 to Infinity of 1/(x^2+1) = Undefined (since Lim x->Infinity of Tan^-1(x) does not exist).

For part b/c. Thanks for anyhelp you can provide in finding my mistake at least for part a so I can recheck part b/c.

 

[slearch]

a) This looks right. You should get the same number as when you look at

-\int_{-\infty}^0 e^x dx

Do you see why?

b)What steps did you go through here?

c) I don't know whether this is undefined, but I do know that

\lim_{x\rightarrow\infty}\arctan (x) = \frac{\pi}{2}

 

[George Jones]

Though, when i do this on my calculator i get something around -.43.

Log on your calculator is log_10, not log_e, and

\int_0^1 log_{10} x dx = -0.43.[/itex]<br /> <br /> In math, the default interpretation of log is log_e; in science, log often means log_10.