Question

How do you evaluate the integral $\int{x{{e}^{-{{x}^{2}}}}}dx$ from $-\infty $ to $\infty $ ?

Answer 1

Hint: We have to compute the value of an integral with indefinite limits that is, from $-\infty $ to $\infty $. Firstly, we will do certain required substitutions in order to simplify our function so that it can be integrated easily. However, since the upper and lower limits of this integral are infinity and negative infinity respectively, thus we will not directly substitute the limits rather we will take the help of summation using limits.

Complete step by step solution:
The function $x{{e}^{-{{x}^{2}}}}$is continuous from $-\infty $ to $\infty $.
Thus, we can break $\int{x{{e}^{-{{x}^{2}}}}}dx$ into two parts as:
$\Rightarrow \int\limits_{-\infty }^{\infty }{x{{e}^{-{{x}^{2}}}}}dx=\int\limits_{0}^{\infty }{x{{e}^{-{{x}^{2}}}}}dx+\int\limits_{-\infty }^{0}{x{{e}^{-{{x}^{2}}}}}dx$
Let $-{{x}^{2}}=t$
Differentiating both sides, we get
$-2x.dx=dt$
$\Rightarrow dx=-\dfrac{dt}{2x}$
 Now we shall substitute the values of $-{{x}^{2}}$ and $dx$ respectively.
$\Rightarrow \int\limits_{-\infty }^{\infty }{x{{e}^{-{{x}^{2}}}}.dx}=\int\limits_{-\infty }^{\infty }{x{{e}^{t}}}\left( -\dfrac{dt}{2x} \right)$
Cancelling x from the right-hand side, we get
$\Rightarrow \int\limits_{-\infty }^{\infty }{x{{e}^{-{{x}^{2}}}}}dx=-\dfrac{1}{2}\left( \int\limits_{0}^{\infty }{{{e}^{t}}dt}+\int\limits_{-\infty }^{0}{{{e}^{t}}}dt \right)$
We know that $\int{{{e}^{x}}.dx={{e}^{x}}+C}$, thus applying the value of this limit, we get
\[\Rightarrow \int\limits_{-\infty }^{\infty }{x{{e}^{-{{x}^{2}}}}.dx}=-\dfrac{1}{2}\left( \left. \left( {{e}^{t}} \right) \right|_{0}^{\infty }+\left. \left( {{e}^{t}} \right) \right|_{-\infty }^{0} \right)\]
We will substitute the value of t back here now.
\[\Rightarrow \int\limits_{-\infty }^{\infty }{x{{e}^{-{{x}^{2}}}}.dx}=-\dfrac{1}{2}\left( \left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{0}^{\infty }+\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{-\infty }^{0} \right)\]
Here, we will solve these two functions separately using limits.
For first function: \[-\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{0}^{\infty }\] ,
\[\begin{align}
  & \Rightarrow -\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{0}^{\infty }=-\dfrac{1}{2}\underset{a\to \infty }{\mathop{\lim }}\,\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{0}^{a} \\
 & \Rightarrow -\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{0}^{\infty }=-\dfrac{1}{2}\left[ \left( {{e}^{-{{a}^{2}}}} \right)-{{e}^{-{{0}^{2}}}} \right] \\
\end{align}\]
\[\Rightarrow -\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{0}^{\infty }=-\dfrac{1}{2}\left[ 0-\left( 1 \right) \right]\]
\[\Rightarrow -\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{0}^{\infty }=\dfrac{1}{2}\] ………………. (1)
For second function: \[-\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{-\infty }^{0}\] ,
\[\begin{align}
  & \Rightarrow -\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{-\infty }^{0}=-\dfrac{1}{2}\underset{b\to -\infty }{\mathop{\lim }}\,\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{b}^{0} \\
 & \Rightarrow -\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{-\infty }^{0}=-\dfrac{1}{2}\left[ \left( {{e}^{-{{0}^{2}}}} \right)-{{e}^{-{{b}^{2}}}} \right] \\
\end{align}\]
\[\Rightarrow -\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{-\infty }^{0}=-\dfrac{1}{2}\left[ 1-\left( 0 \right) \right]\]
\[\Rightarrow -\dfrac{1}{2}\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{-\infty }^{0}=-\dfrac{1}{2}\] ………………. (2)
Combining (1) and (2), we get
\[\begin{align}
  & \Rightarrow \int\limits_{-\infty }^{\infty }{x{{e}^{-{{x}^{2}}}}.dx}=-\dfrac{1}{2}\left( \left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{0}^{\infty }+\left. \left( {{e}^{-{{x}^{2}}}} \right) \right|_{-\infty }^{0} \right) \\
 & \Rightarrow \int\limits_{-\infty }^{\infty }{x{{e}^{-{{x}^{2}}}}.dx}=\dfrac{1}{2}-\dfrac{1}{2} \\
 & \Rightarrow \int\limits_{-\infty }^{\infty }{x{{e}^{-{{x}^{2}}}}.dx}=0 \\
\end{align}\]
Therefore, the value of the integral $\int{x{{e}^{-{{x}^{2}}}}}dx$ from $-\infty $ to $\infty $ is 0.

Note: While doing the calculations after applying the value of limit, one must be very careful about putting the correct values of limit as the terms on the right hand side get more complex. The possible mistake we could have made was that we could have forgotten to finally re-substitute the value of our assumed t-variable or we could have proceeded further to calculate the integral without using the limits.