Evaluate $$\int_2^\infty\frac{\ln(t-1)}{(t-1)^5}\,dt.$$
I think I have come pretty close to an answer on this one by using substitution and integration by parts.
However, I'm pretty unsure how to reason in the last step when inserting infinity into the antiderivative. One expression becomes $-\infty/\infty$, should I evaluate that expression by using L'Hospital's rule or does anyone have any suggestions?
Thanks in advance for helping out!
In your substitution $u=t-1$ you forgot to adjust the limits. So we have $$\int_2^\infty\frac{\ln(t-1)}{(t-1)^5}\,dt=\int_1^\infty\frac{\ln u}{u^5}\,du.$$ Integration by parts with \begin{align}f(u)=\ln u&\implies f'(u)=1/u\\g'(u)=1/u^5&\implies g(u)=-1/(4u^4)\end{align} yields \begin{align}\int_1^\infty\frac{\ln u}{u^5}\,du&=\left[-\frac{\ln u}{4u^4}\right]_1^\infty-\int_1^\infty\frac1u\left(-\frac1{4u^4}\right)\,du\\&=-\lim_{u\to\infty}\frac{\ln u}{4u^4}+\frac{\ln1}{4\cdot1^4}+\frac14\int_1^\infty\frac{du}{u^5}\\&=-\lim_{u\to\infty}\frac{(\ln u)'}{(4u^4)'}+0+\frac14\left[-\frac1{4u^4}\right]_1^\infty\tag{L'Hopital}\\&=-\lim_{u\to\infty}\frac{1/u}{16u^3}+\frac14\left[-0-\left(-\frac1{4\cdot1^4}\right)\right]\\&=-\frac1{16}\lim_{u\to\infty}\frac1{u^4}+\frac14\cdot\frac14\\&=\frac1{16}.\end{align} There is also no need to convert $u$ back into $t-1$ after the integration as the integral is definite.