Determine whether the improper integral converges or diverges.
-$$\int _ { 1 } ^ { \infty } \frac { x ^ { 6 } } { \sqrt { e ^ { x } - 1 } } d x$$
A) Diverges
B) Converges

Question

Quizplus · Accepted Answer

We can use the limit comparison test with $\frac{1}{\sqrt{x}}$ to see that the improper integral converges.

First, note that $\frac{x^6}{\sqrt{e^x-1}} > x^6$ for all $x \geq 1$, since $e^x > 1$.

Then, we have:

$$\lim_{x 	o \infty} \frac{x^6}{x^{3/2}} = \lim_{x 	o \infty} x^{9/2} \cdot \frac{1}{x^{3/2}} = \lim_{x 	o \infty} x^3 = \infty$$

Since $\frac{x^6}{\sqrt{e^x-1}}$ is greater than $x^6$ for large $x$, and $\int_1^\infty x^6 dx$ converges (since $6 > 1$), we can use the limit comparison test to say that $\int_1^\infty \frac{x^6}{\sqrt{e^x-1}} dx$ also converges.

Determine Whether the Improper Integral Converges or Diverges ∫1∞x6ex−1dx\int _ { 1 } ^ { \infty } \frac { x ^ { 6 } } { \sqrt { e ^ { x } - 1 } } d x∫1∞​ex−1​x6​dx

Determine Whether the Improper Integral Converges or Diverges $\int _ { 1 } ^ { \infty } \frac { x ^ { 6 } } { \sqrt { e ^ { x } - 1 } } d x$