Computing $\int_1^\infty\frac{\ln(x)}{x^2}\,dx.$

Question

I'm computing $$\int_1^\infty\frac{\ln(x)}{x^2}\,dx.$$

I have the proper antiderivative.

My first strategy in finding the limit involved splitting the limit into two, resulting in:

lim t>inf (-ln(x)/x) - lim t>inf (1/x), evaluated from 1 to t

I then used L'Hospital's rule to eliminate inf/inf in the first limit, and then evaluated each from 1 to t.

Using this method, I got 2.

Then I did the same thing, but kept the fraction intact: -(ln(x) + 1)/x. Using L'HR here and then evaluating from 1 to t, I got 1 (which is the answer all the calculators give me).

What am I missing with the first method? Did I not use L'HR properly (only on one of the limits). Obviously, the second method gets rid of the 1/x when using L'HR, but I can't find something that tells me I can't do that. Even my professor is stumped right now.

Thanks

Answer 1

Notice that $$\int_1^\infty \frac{\log x}{x^2}dx = \lim_{t\to\infty}\int_1^t \frac{\log x}{x^2}dx = \lim_{t\to\infty}\left.-\frac{\log x}{x}\right\rvert^t_1+\int_1^\infty\frac{dx}{x^2}= \lim_{t\to\infty}\left.-\frac{\log x}{x}\right\rvert^t_1\left.-\frac1x\right\rvert^t_1.$$ Now, both limits exist independently and \begin{aligned} \lim_{t\to\infty}\frac{\log t}{t} = \lim_{t\to\infty}\frac 1t = 0. \end{aligned} Therefore, since $\log 1 = 0$, $$\int_1^\infty \frac{\log x}{x^2}dx = 1.$$

Answer 2

For $a\gt1$, $$ \begin{align} \int_1^\infty\frac{\log(x)}{x^a}\,\mathrm{d}x &=-\frac{\mathrm{d}}{\mathrm{d}a}\int_1^\infty x^{-a}\,\mathrm{d}x\tag1\\ &=-\frac{\mathrm{d}}{\mathrm{d}a}\frac1{a-1}\tag2\\ &=\frac1{(a-1)^2}\tag3 \end{align} $$ Explanation:
$(1)$: $-\frac{\mathrm{d}}{\mathrm{d}a}x^{-a}=\frac{\log(x)}{x^a}$
$(2)$: integrate
$(3)$: differentiate

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A question was raised about the interchange of the integral and the derivative in step $(1)$. We can justify that with Fubini-Tonelli and the Fundamental Theorem of Calculus: $$ \begin{align} -\frac{\mathrm{d}}{\mathrm{d}a}\int_1^\infty\frac{\mathrm{d}x}{x^a} &=-\frac{\mathrm{d}}{\mathrm{d}a}\int_1^\infty\int_a^\infty\frac{\log(x)}{x^t}\,\mathrm{d}t\,\mathrm{d}x\\ &=-\frac{\mathrm{d}}{\mathrm{d}a}\int_a^\infty\int_1^\infty\frac{\log(x)}{x^t}\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_1^\infty\frac{\log(x)}{x^a}\,\mathrm{d}x \end{align} $$

Answer 3

Substitute $\ln(x)=t$ to get $\displaystyle \int_{1}^{\infty}\frac{\ln(x)}{x^{2}}dx=\int_{0}^{\infty}te^{-t}dt=\Gamma(2)=1$