$$ \lim_{R \rightarrow \infty} \int_{-R}^{R} \frac{x}{1+x^3} \ dx$$
It is allowed to use Taylor series if that helps in any way.
Asked
Active
Viewed 57 times
-3
-
silly question, but there is no differential. And can $x$ be complex? – imranfat Dec 12 '20 at 07:17
-
x is real. @RagibZaman I have not worked with contours before,. – Labbsserts Dec 12 '20 at 07:28
-
@imranfat It is with respect to x – Labbsserts Dec 12 '20 at 07:43
-
2As long as $R>1$, the integral $\int_{-R}^{R}\frac{x}{1+x^3},\mathrm{d}x$ suffers from the singularity at $x=-1$, rendering it divergent. So the limit is also divergent. – Sangchul Lee Dec 12 '20 at 08:42
1 Answers
1
Using sum of cubes we have $x^3+1 = (x+1)(x^2-x+1)$. This can be used to find the partial fraction decomposition $$\frac{x}{x^3+1} = \frac{x+1}{3(x^2-x+1)}-\frac{1}{3(x+1)}.$$ Now, the first term on the right has finite positive integral on the interval $(-1,0)$ (it is not too hard to check that it is nonnegative and bounded). The second term, however, has divergent integral on this interval. Indeed, with $u=x+1$, we have $$\int_{-1}^0 \frac{dx}{x+1} = \int_0^1\frac{du}{u},$$ and it is a standard exercise that the latter is infinite (compute proper integrals on $(\epsilon, 1)$ as $\epsilon\to 0^+$ to see this).
Since our original function is the difference of something with finite positive integral and divergent integral on $(-1,0)$, it has divergent integral there as well. Thus if $R>1$ the integral inside your limit diverges.
Glare
- 3,639