Calculus Question: $\int_2^\infty\frac{\log^3(x-1)}{x^2}dx$ I have just taken calculus quiz but I could not find  $\displaystyle \int_2^\infty\frac{\log^3(x-1)}{x^2}dx$? Any help would be appreciated. Thanks in advance.
EDIT:
Forgot to mention, my tutor gave us hints about this question.


*

*Use Taylor series

*$\displaystyle \zeta(3)=\sum_{n=1}^{\infty}\frac{1}{n^3}$


Those the hints that she gave to us.

Use the substitution $x-1=t$ to obtain:
$$\int_1^{\infty} \frac{\log^3t}{(1+t)^2}\,dt$$
With the substitution $t=1/u$, the above integral is:
$$\int_0^1 \frac{-\log^3 u}{(1+u)^2}\,du$$
Next, use the following series representation,
$$\frac{1}{(1+u)^2}=\sum_{n=0}^{\infty}(-1)^n (n+1)u^n$$
to obtain:
$$\int_0^1 \frac{-\log^3 u}{(1+u)^2}\,du=-\sum_{n=0}^{\infty} (-1)^n(n+1)\int_0^1 \log^3u \,u^n\,du$$
Next use the substitution $\log u=-x$ to get:
$$\sum_{n=0}^{\infty} (-1)^n(n+1)\int_0^{\infty} x^3e^{-(n+1)x}\,dx$$
With another substitution $(n+1)x=y$, you should get:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^3}\int_0^{\infty} y^3e^{-y}\,dy$$
Recoginse that $\int_0^{\infty} y^3e^{-y}\,dy=\Gamma(4)=3!$ (you can show this using integration by parts if you don't like the Gamma function), hence
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^3}\int_0^{\infty} y^3e^{-y}\,dy=6\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^3}$$
It is easy to show that:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^3}=\frac{3\zeta(3)}{4}$$
Hence, our final answer is:
$$\boxed{\dfrac{9\zeta(3)}{2}}$$
