Problem 42 - Calculus Early Transcendentals

Question

Evaluate the following integrals or state that they diverge. $$\int_{0}^{\ln 3} \frac{e^{y}}{\left(e^{y}-1\right)^{2 / 3}} d y$$

Step-by-Step Solution

Answer

Question: Evaluate the integral or determine if it is divergent: $$\int_{0}^{\ln 3} \frac{e^{y}}{\left(e^{y}-1\right)^{2 / 3}} d y$$ Answer: The given integral converges and has a value of $$3(2)^{1/3}$$.

Step 1: Identify any singularities in the integral.

The given integral is: $$\int_{0}^{\ln 3} \frac{e^{y}}{\left(e^{y}-1\right)^{2 / 3}} d y$$ Note that when y=0, we have the denominator $$\left(e^{0}-1\right)^{2 / 3} = 0$$, which results in a potential singularity.

Step 2: Divide the integral at the singularity.

In order to evaluate the integral, we will divide it into two parts, one approaching the singularity from the left and the other from the right. This gives us: $$\lim_{a \to 0^{+}} \int_{a}^{\ln 3} \frac{e^{y}}{\left(e^{y}-1\right)^{2 / 3}} d y$$

Step 3: Find an antiderivative for the integrand.

We will find an antiderivative for the integrand using substitution. Let: $$u = e^{y} - 1$$ Then, the differential form for u is: $$du = e^{y} dy$$ Now, we can rewrite the integral in terms of u: $$\int \frac{e^{y}}{\left(e^{y}-1\right)^{2 / 3}} dy = \int \frac{1}{(u)^{2/3}} du$$ Now, we can integrate the new integrand with respect to u. We obtain: $$\int \frac{1}{(u)^{2/3}} du = \int u^{-2/3} du = \frac{3}{1}u^{1/3} + C = 3(u)^{1/3} + C$$ Now, we convert back to y: $$3(\left(e^{y} - 1\right)^{1/3}) + C$$

Step 4: Evaluate the integral using the Fundamental Theorem of Calculus.

Now we can use the antiderivative and the Fundamental Theorem of Calculus to evaluate the limit $$\lim_{a \to 0^{+}} \int_{a}^{\ln 3} \frac{e^{y}}{\left(e^{y}-1\right)^{2 / 3}} d y = \lim_{a \to 0^{+}} \left[\left. 3(\left(e^{y} - 1\right)^{1/3})\right|_{a}^{\ln 3} \right]$$

Step 5: Evaluate the limit to determine convergence or divergence.

We have: $$\lim_{a \to 0^{+}} \left(3(\left(e^{\ln(3)} - 1\right)^{1/3} - 3(\left(e^{a} - 1\right)^{1/3})\right)$$ As a approaches 0 from the positive direction, $$\left(e^{a}-1\right)^{1/3}$$ approaches 0. Therefore, the limit converges to: $$3(\left(3 - 1\right)^{1/3}) - 3\cdot 0 = 3(2)^{1/3}$$ Hence, the value of the given integral is: $$\int_{0}^{\ln 3} \frac{e^{y}}{\left(e^{y}-1\right)^{2 / 3}} d y = 3(2)^{1/3}$$