Question

7–58. Improper integrals Evaluate the following integrals or state that they diverge.25. ∫ (from -∞ to ∞) e³ˣ/(1 + e⁶ˣ) dx

Accepted Answer

First, recognize that the integral is an improper integral with limits from $$-\infty to$$ $$\infty$$, so we need to consider the behavior of the integrand as $$x 	o -\infty$$ and $$x 	o \infty. $$Rewrite the integrand to see if it can be simplified. The integral is $$\int_{-\infty}^{\infty} \frac{e$$^{3x}$$}{1 + e$$^{6x}$$} \, dx. $$Notice that $$e^{6x}$$ = $$(e^{3x})^2$, so the denominator is 1$$ + $$(e^{3x})^2$. Make the substitution t$$ = $$e^{3x}. $$Then, $$dt = 3 e$$^{3x}$$ dx = 3 t dx$$, which implies $$dx = \frac{dt}{3t}. $$Also, as $$x 	o -\infty$$, $$t 	o 0$$, and as $$x 	o \infty$$, $$t 	o \infty. $$Rewrite the integral in terms of $$t$$: $$\int_{0}^{\infty} \frac{t}{1 + t^2$} \cdot \frac{1}{3t} dt = \frac{1}{3} \int_{0}^{\infty} \frac{1}{1 + t^2$} dt. $$Notice how the $$t in $$numerator and denominator cancel out. Recognize that $$\int \frac{1}{1 + t^2$} dt is $$the standard integral for $$\arctan t. So$$, the integral becomes $$\frac{1}{3} [\arctan t]_{0}^{\infty}. $$Evaluate this expression to determine the value of the integral or conclude if it converges.

7–58. Improper integrals Evaluate the following integrals or state that they diverge.
25. ∫ (from -∞ to ∞) e³ˣ/(1 + e⁶ˣ) dx

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Improper Integrals

Substitution Method

Behavior of Exponential Functions