Calculating improper integral limits

Question

I am trying to calculate $\int_{-\infty }^1 {dx\over x^{1/3}}$. I have come up with $\int_{-\infty }^1 {dx\over x^(1/3)}$=$\int_{-\infty }^0 {dx\over x^{1/3}}$+$\int_{0}^1 {dx\over x^{1/3}}$.

Then I am trying to calculate $\int_{-\infty }^0 {dx\over x^(1/3)}$ and for that I am calculaing first $\int_{a}^0 {dx\over x^{1/3}}=(-3/2)a^{ (2/3)}$.

Then I am trying to calculate $$\lim_{a\to-\infty} (-3/2)a^{ (2/3)}$$

At this point my best guess is that the above limit does not exist but I am not sure. Can somebody please verify if the limit exists or not?

replace "^(1/3)" through "^{1/3}" in the tex code please – tired Jun 25 '15 at 06:35 — tired, Jun 25 '15 at 06:35
btw, yes this integral diverges at the lower bound – tired Jun 25 '15 at 06:36 — tired, Jun 25 '15 at 06:36

score 0 · Accepted Answer · answered Jun 25 '15 at 06:37

0

$\displaystyle \int_{-\infty}^1 \dfrac{1}{x^{\frac{1}{3}}}dx=\displaystyle \int_{-\infty}^1 x^{-\frac{1}{3}}dx=\dfrac{3}{2}x^{\frac{2}{3}}|_{-\infty}^1=-\infty$

answered Jun 25 '15 at 06:37

DeepSea

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Calculating improper integral limits

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