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

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In summary, the conversation discusses using integration by parts to evaluate the improper integral $\int_{1}^{\infty} {x}^{-3} lnx \,dx$, which simplifies to $\frac{1}{4}$. The solution is confirmed using a substitution and integration by parts.
  • #1
tmt1
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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.
 
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  • #2
This is how I would work it. We are given:

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

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

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

Using IBP, where:

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

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

And so we have:

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

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

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

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

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

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

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

\(\displaystyle I=\frac{1}{4}\)
 
  • #3
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.
 

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

What is the formula for integrating $\int_{1}^{\infty} {x}^{-3} lnx \,dx$?

The formula for integrating $\int_{1}^{\infty} {x}^{-3} lnx \,dx$ is $\lim_{b \to \infty} \int_{1}^{b} {x}^{-3} lnx \,dx$.

What is the value of the integral $\int_{1}^{\infty} {x}^{-3} lnx \,dx$?

The value of the integral $\int_{1}^{\infty} {x}^{-3} lnx \,dx$ is undefined as the function diverges at infinity.

Is the integral $\int_{1}^{\infty} {x}^{-3} lnx \,dx$ convergent or divergent?

The integral $\int_{1}^{\infty} {x}^{-3} lnx \,dx$ is divergent as the function goes to infinity.

Can the integral $\int_{1}^{\infty} {x}^{-3} lnx \,dx$ be solved using substitution?

No, substitution is not a viable method for solving this integral as the function does not have a closed form antiderivative.

How can the integral $\int_{1}^{\infty} {x}^{-3} lnx \,dx$ be approximated?

The integral $\int_{1}^{\infty} {x}^{-3} lnx \,dx$ can be approximated using numerical integration methods such as Simpson's rule or trapezoidal rule.

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