Integrating $\int_{1}^{\infty} {x}^{-3} lnx \,dx$: Is the Answer 1?

[tmt1]

I have

$$\int_{1}^{\infty} {x}^{-3} lnx \,dx$$

I choose to use integration by parts so I let $u = lnx$ and $dv = {x}^{-3} dx$.

Therefore $du = \frac{1}{x} dx$ and $v = \frac{{x}^{-2}}{-2}$

Thus, what I need to evaluate is

$$- \frac{-lnx}{{2x}^{2}} - \frac{1}{4 {x}^{2}} + C$$

As $x$ approaches infinity, both those terms go to zero. So I am left with:

$$\frac{-ln1}{{2(1)}^{2}} + \frac{1}{4 ({1}^{2})} $$

Since $ln1 = 0$ the first term evaluates to zero, so I am left with $+ \frac{1}{4}$.

Is this correct? The solution is saying it evaluates to 1.

 

[MarkFL]

This is how I would work it. We are given:

$$I=\int_1^{\infty} x^{-3}\ln(x)\,dx$$

This is an improper integral, so let's use a limit:

$$I=\lim_{t\to\infty}\left(\int_1^{t} x^{-3}\ln(x)\,dx\right)$$

Using IBP, where:

$$u=\ln(x)\,\therefore\,du=x^{-1}\,dx$$

$$dv=x^{-3}\,dx\,\therefore\,v=-\frac{1}{2}x^{-2}$$

And so we have:

$$I=\lim_{t\to\infty}\left(\left[-\frac{1}{2}x^{-2}\ln(x)\right]_1^t+\frac{1}{2}\int_1^t x^{-3}\,dx\right)$$

$$I=\lim_{t\to\infty}\left(-\frac{1}{2}t^{-2}\ln(t)-\frac{1}{4}\left[x^{-2}\right]_1^t\right)$$

$$I=\lim_{t\to\infty}\left(-\frac{1}{2}t^{-2}\ln(t)-\frac{1}{4}\left(t^{-2}-1\right)\right)$$

$$I=\lim_{t\to\infty}\left(-\frac{1}{2}t^{-2}\ln(t)-\frac{1}{4}t^{-2}+\frac{1}{4}\right)$$

$$I=\frac{1}{4}-\frac{1}{2}\lim_{t\to\infty}\left(\frac{\ln(t)}{t^2}\right)$$

Now, application of L'Hôpital's Rule gives us:

$$I=\frac{1}{4}-\frac{1}{2}\lim_{t\to\infty}\left(\frac{1}{2t^2}\right)$$

$$I=\frac{1}{4}$$

 

[alyafey22]

The integral can be written as

$$-\int^\infty_1 (1/x)\frac{\ln(1/x)}{x^2} dx$$

use the sub $1/x = t$

$$-\int^1_0t\ln(t) dt$$

finalize by integration by parts.