Determine whether the improper integral $\int _ { - \infty } ^ { \infty } \frac { 9 } { 49 + x ^ { 2 } } d x$ diverges or converges. Evaluate the integral if it converges.
A) $\frac { 9 } { 49 }$
B) $\frac { 9 } { 7 } \pi$
C) $\frac { 9 } { 49 } \pi$
D) $\frac { 9 } { 7 }$
E) diverges

Question

Quizplus · Accepted Answer

To determine whether the improper integral converges or diverges, we need to evaluate the integral in terms of a limit. We can do this by splitting the integral into two parts: 
$$\int_{-\infty}^{\infty} \frac{9}{49+x^2} dx = \lim_{a\to-\infty} \int_{a}^{c} \frac{9}{49+x^2} dx + \lim_{b\to\infty} \int_{c}^{b} \frac{9}{49+x^2} dx$$
where $c$ is any constant between $-\infty$ and $\infty$.

Using the substitution $x=\sqrt{49}\tan\theta$, we can evaluate the integral: 
$$ \int \frac{9}{49+x^2} dx = \int \frac{\frac{9}{49}}{1+(\frac{x}{7})^2} dx=9\int \frac{1}{1+(\frac{x}{7})^2} dx = 9\cdot7\tan^{-1}(\frac{x}{7})+C $$

Therefore, the improper integral becomes: 
$$\lim_{a\to-\infty} [9\cdot7\tan^{-1}(\frac{x}{7})]_{a}^{c} + \lim_{b\to\infty} [9\cdot7\tan^{-1}(\frac{x}{7})]_{c}^{b}$$
Since $\tan^{-1}$ is bounded between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, we have: 
$$\lim_{a\to-\infty} [9\cdot7\tan^{-1}(\frac{x}{7})]_{a}^{c} = 9\cdot7(\frac{\pi}{2}+\frac{\pi}{2}) = 63\pi$$
$$\lim_{b\to\infty} [9\cdot7\tan^{-1}(\frac{x}{7})]_{c}^{b} = 9\cdot7(0-0) = 0$$
Therefore, the integral converges, and its value is 
$$\int_{-\infty}^{\infty} \frac{9}{49+x^2} dx = 63\pi + 0 = \boxed{\textbf{(B) } \frac{9}{7}\pi}$$

Determine Whether the Improper Integral ∫−∞∞949+x2dx\int _ { - \infty } ^ { \infty } \frac { 9 } { 49 + x ^ { 2 } } d x∫−∞∞​49+x29​dx

Determine Whether the Improper Integral $\int _ { - \infty } ^ { \infty } \frac { 9 } { 49 + x ^ { 2 } } d x$