Question

Evaluate ${\int }_{-\infty }^{\infty }x{e}^{-x^2}dx$

• A. $\frac{1}{2}$
• B. $\frac{-1}{2}$
• C. $0$
• D. None of the above

Answer 1

Answer: (C)

$0$

Given : ${\int }_{-\infty }^{\infty }x{e}^{-x^2}dx$

${\int }_{-\infty }^{\infty }x{e}^{-x^2}dx=\stackrel{lim}{c\to -\infty }{\int }_c^{0}x{e}^{-x^2}dx+\stackrel{lim}{b\to \infty }{\int }_0^{b}x{e}^{-x^2}=\stackrel{lim}{c\to -\infty }{\int }_c^{0}x{e}^{-x^2}dx+\stackrel{lim}{b\to \infty }{\int }_0^b(-\frac{1}{2})(-2x)e^{-x^2}dx$

Let $z=-x^2$

$x\to 0dz=-2xdx$

${\int }_{-\infty }^{\infty }x{e}^{-x^2}dx=\stackrel{lim}{c\to -\infty }{\int }_c^0(-\frac{1}{2})e^{z}dz+\stackrel{lim}{b\to \infty }{\int }_0^b(-\frac{1}{2})e^{z}dz=\stackrel{lim}{c\to -\infty }{[-\frac{e^z}{2}]}_c^0+\stackrel{lim}{b\to \infty }{[-\frac{e^z}{2}]}_0^b=\stackrel{lim}{c\to -\infty }[+\frac{e^{-c^2}}{2}-\frac{e^0}{2}]+\stackrel{lim}{b\to \infty }[-\frac{e^{-b^2}}{2}+\frac{e^0}{2}]=-\frac{1}{2}+\frac{1}{2}=0$

Hence the correct answer is $0$.