# Integration by Parts Problem: Help in understanding why a part of it equals 0

$$4I= \int_0^{\infty} \frac{4x^3 +\sin(3x)-3\sin x}{x^5} \ \mathrm{d}x$$ $$=\frac{-1}{4} \underbrace{\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}}_{=0} +\frac{1}{4} \int_0^{\infty} \frac{12x^2 -3\cos x +3\cos(3x)}{x^4} \ \mathrm{d}x$$

Now, I was told that the brackets gives zero, but not why it is so. Could somebody please help explain this to me? Many thanks!

• What are the limits as that term tends to $\infty$ and 0? Jun 1, 2015 at 15:28
• Sorry, I couldn't understand. How do you evaluate the bracket at $x=0$? Jun 1, 2015 at 15:33
• Probably not the most elegant way to do this but the taylor expansion of that term around $x=0$ is given by $2x-\frac{13x^3}{30} + \mathcal{O}(x^5)$. Jun 1, 2015 at 15:43
• The most direct thing to do with $0$ is to treat it as an improper integral. It turns out you can just continuously extend the integrand, though.
– user14972
Jun 1, 2015 at 16:05

$${\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}} =$$ $$\lim_{x\to \infty}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )-\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$ take a look on $$\lim_{x\to \infty}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$ $sin(3x)$ and $3sin(x)$ are a bounded functions and the deg($4x^3$)< deg($x^4$). Then you'll find that the value of this limit, when x goes to infinity, is cero.
What about, $$\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )$$ It's obvious that the functions $f(x)=4x^3+\sin(3x)- 3 \sin x$ and $g(x)={x^4}$ cab be derived four times so if we apply L'Hopital theorem 4 times we'll find that $$\lim_{x\to 0}(\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} )=\lim_{x\to 0}(\frac{\ 81sin(3x)- 3 \ sinx}{24} )=0$$
In conclusion $${\left[\frac{4x^3+\sin(3x)- 3 \sin x}{x^4} \right]_0^{\infty}} =0-0=0$$