(disclaimer: I don't know if it's legit to share a picture of my calculations, and not rewriting it in LATEX notation – let me know if this is legit or not)
my question is that: are my calculations correct?
I have to solve this exercise: it's an improper integral.
before using integration by parts, I've studied the bounds, in order to check where the function is undefined. And I got that when solving the function for $x = 3$ the denominator is 0, therefore the function is undefined (i.e division by zero).
Therefore, I've taken the limit as x approaches 3. And then I've solved the integral with $x$ as the upper bound.
As stated above, I've used integration by parts by choosing $$ u = x $$ (because the derivative of the polynomial, hopefully, is going to become some smaller value) And $$ dv = (1 / (3 – x)) $$ because the antiderivative is simply equal to log (natural log, i.e with base $e$) (of course the argument of log must be the absolute value)
Now, by integrating by parts and after having evaluated the limit of the antiderivative, I got that the limit doesn't exist, because the limit of the function evaluated in the upper bound is undefined (i.e the natural log is undefined for x = 3). Is it true? And if it's true, the integral doesn't converge, right?
Best Answer
You are correct that the integral does not converge, but you made some mistakes and overcomplicated the solution in general.
The mistake: If $v=\log(3-x)$, then $v'=-\frac{1}{3-x}$. You missed a minus sign.
The overcomplication:
Instead of using per partes, you can rewrite
$$\frac{x}{3-x} = \frac{x-3+3}{3-x} = \frac{-(3-x)}{3-x} + \frac{3}{3-x} = \frac{3}{3-x} - 1$$
and only integrate after this rearrangement. No need for per partes, a simple introduction of a new variable $u=3-x$ is sufficient and you get (since $du = -dx$):
$$\int_1^3\frac{x}{3-x}dx = 3\int_1^3 \frac{1}{3-x}dx - \int_1^3 1dx = 3\int_2^0-\frac{1}{u}du - 2 = 3\int_0^2\frac1udu - 2$$
now you can either remember that the integral of $\frac{1}{u}$ diverges around $0$, or you can write it out, since
$$\int_0^2\frac1udu=\lim_{x\to 0}\int_x^2\frac1udu = \lim_{x\to 0} (\ln(2)-\ln(x))$$ and the limit above does not exist.