Integration of $\int_0^4 \frac{1}{\sqrt{|x-3|}} dx$

Question

I believe it might diverge, but how do you evaluate/show that it diverges (if it does)?

I am especially not sure about how to deal with the absolute value sign when integrating.

Answer 1

Notice that the x-value of $3$ is in [0,4] and makes the denominator of the inegrand 0. You'll need to split this up into 2 improper integrals. $$\int_0^4 \frac{dx}{\sqrt{|x-3|}} =\int_0^{3^-} \frac{dx}{\sqrt{|x-3|}}+\int_{3^+}^4 \frac{dx}{\sqrt{|x-3|}}$$

Each integral you'll want to handle with a one-sided limit:

$$\int_0^{3^-} \frac{dx}{\sqrt{|x-3|}}=\lim_{b\to 3^-}\int_0^{b} \frac{dx}{\sqrt{|x-3|}}$$

and

$$\int_{3^+}^{4} \frac{dx}{\sqrt{|x-3|}}=\lim_{a\to 3^+}\int_a^{4} \frac{dx}{\sqrt{|x-3|}}$$

I'll leave the rest to you :)