improper integral

**stumped765** · November 15th 2009, 01:09 PM

show the improper integral is well defined.

$\int_0^{\infty} \frac{x^{2/3}}{e^{2x}-1}\,dx$

**Opalg** · November 16th 2009, 12:32 AM

Originally Posted by stumped765

show the improper integral is well defined.

$\int_0^{\infty} \frac{x^{2/3}}{e^{2x}-1}\,dx$

Split the integral into two, $\textstyle\int_0^\infty = \int_0^1+\int_1^\infty$ , then use the comparison test.

Near x=0, $e^{2x}\approx 1+2x$ . So $\frac{x^{2/3}}{e^{2x}-1}\approx \tfrac12x^{-1/3}$ , and $\int_0^1x^{-1/3}dx$ converges.

For large x, $e^{2x}-1>x^2$ (because exponentials grow faster than any power of x). So $\frac{x^{2/3}}{e^{2x}-1} < x^{-4/3}$ , and $\int_1^\infty x^{-4/3}dx$ converges.

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