Solve the problem.
-Evaluate
$$\int _ { 0 } ^ { \infty } \frac { e ^ { - 3 x ^ { 2 } } - e ^ { - 6 x ^ { 2 } } } { x } d x$$
by writing the integrand as an integral.
A) $\frac { 1 } { 2 } \ln 2$
B) $\frac { 1 } { 2 } \ln \frac { 1 } { 2 }$
C) $\ln \frac { 1 } { 2 }$
D) $\ln 2$

Question

Quizplus · Accepted Answer

We can write the integrand as $$\int_0^\infty e^{-3x^2}\left(\frac{1}{x}-\frac{1}{\sqrt{2}x}\right)e^{(3-6)x^2}dx.$$ Notice that $\int_0^\infty e^{(3-6)x^2}dx$ is just a normalizing constant for a Gaussian distribution, and evaluates to $\frac{\sqrt{\pi}}{2\sqrt{3}}.$ Thus, the original integral becomes $$\int_0^\infty e^{-3x^2}\left(\frac{1}{x}-\frac{1}{\sqrt{2}x}\right)\cdot\frac{\sqrt{\pi}}{2\sqrt{3}}dx.$$ Using the substitution $u=\sqrt{3}x$ and then splitting the integrand, we have $$\int_0^\infty e^{-3x^2}\cdot\frac{1}{x}\cdot\frac{\sqrt{\pi}}{2\sqrt{3}}dx-\frac{1}{\sqrt{2}}\int_0^\infty e^{-3x^2}\cdot\frac{1}{u}\cdot\frac{\sqrt{\pi}}{2\sqrt{3}}du.$$ Evaluating the first integral gives $\frac{1}{2}\ln 3,$ and evaluating the second integral gives $\frac{1}{2}\ln 2.$ Thus, the total value of the integral is $\boxed{\frac{1}{2}\ln 2}.$

Solve the Problem ∫0∞e−3x2−e−6x2xdx\int _ { 0 } ^ { \infty } \frac { e ^ { - 3 x ^ { 2 } } - e ^ { - 6 x ^ { 2 } } } { x } d x∫0∞​xe−3x2−e−6x2​dx

Solve the Problem $\int _ { 0 } ^ { \infty } \frac { e ^ { - 3 x ^ { 2 } } - e ^ { - 6 x ^ { 2 } } } { x } d x$