MHB What Am I Doing Wrong? : Improper Intergals

[Pull and Twist]

Here's what I am trying to evaluate...

$$\int_{0}^{2}z^2lnz\,dz$$

This is what I did...

$$\lim_{{t}\to{0}}\int_{t}^{2}z^2lnz \,dz$$

Then using integration by parts...

$$u=\ln\left({z}\right)$$

$$du=\frac{1}{z}dz$$

$$dv=z^2dz$$

$$v=\frac{z^3}{3}$$

$$\lim_{{t}\to{0}}\left[\frac{z^3}{3}\ln\left({z}\right)\right]-\frac{1}{3}\int_{t}^{2}z^2\,ds$$

$$\lim_{{t}\to{0}}\frac{1}{3}\left[z^3\ln\left({z}\right)\right]-\frac{1}{9}\left[z^3\right]$$ for $$t \le z\le2$$

I know that $$\left[z^3\ln\left({z}\right)\right]$$ will give me some trouble so I use L'hopital's rule.

$$\left[\frac{\ln\left({z}\right)}{z^\left(-3\right)}\right]\implies

\left[\frac{\frac{1}{z}}{-3z^\left(-4\right)}\right]\implies

-\frac{1}{3}\left[z^3\right]$$

Then I plug that all back into my equation...

$$\lim_{{t}\to{0}}-\frac{1}{9}\left[z^3\right]-\frac{1}{9}\left[z^3\right]$$ for $$t \le z\le2$$

After evaluating I get...

$$-\frac{1}{9}\left[8-0\right]-\frac{1}{9}\left[8-0\right]\implies-\frac{16}{9}$$

What am I doing wrong? The solution manual states that I should be getting $$\frac{8}{3}\ln\left({2}\right)-\frac{8}{9}$$ as my answer.

 

[Pull and Twist]

Oops, I think I see where I went wrong... I should only be using L'hopital's rule for the portion of $$\left[z^3\ln\left({z}\right)\right]$$ evaluated at $$t=0$$

In which case I would get $$-\frac{1}{3}\left[z^3\right]\implies-\frac{1}{3}\left[0^3\right] =0$$

So evaluated at $$\frac{1}{3}\left[2^3\ln\left({2}\right) - 0\right]-\frac{8}{9}$$ I would get the correct answer.

Lol. Sometimes all it takes is typing it up on here.

 

[MarkFL]

After you apply IBP, you should have:

$$\frac{1}{3}\lim_{t\to0^{+}}\left[\left[z^3\ln(z)\right]_t^2-\int_t^2 z^2\,dz\right]$$

$$\frac{1}{3}\lim_{t\to0^{+}}\left[8\ln(2)-t^3\ln(t)-\frac{1}{3}\left[z^3\right]_t^2\right]$$

$$\frac{1}{3}\lim_{t\to0^{+}}\left[8\ln(2)-t^3\ln(t)-\frac{1}{3}\left(8-t^3\right)\right]$$

$$\frac{8}{3}\ln(2)-\frac{8}{9}-\frac{1}{3}\lim_{t\to0^{+}}\left[\frac{\ln(t)}{t^{-3}}\right]$$

Now apply L'Hópital's Rule:

$$\frac{8}{3}\ln(2)-\frac{8}{9}+\frac{1}{9}\lim_{t\to0^{+}}\left[t^3\right]$$

And your final answer is:

$$\frac{8}{3}\ln(2)-\frac{8}{9}=\frac{8}{9}\left(3\ln(2)-1\right)$$

 

[Pull and Twist]

Thanks Mark. I shortly figured out where I went wrong after posting this, but your explanation was insightful none the less.