Video Transcript
The integral between negative
infinity and positive infinity of 𝑥 multiplied by 𝑒 to the power of negative 𝑥
squared with respect to 𝑥 is convergent. What does it converge to?
We have a formula which can help us
to evaluate integrals of this form. The formula tells us that the
integral from negative infinity to positive infinity of 𝑓 of 𝑥 with respect to 𝑥
is equal to the limit as 𝑠 tends to negative infinity of the integral from 𝑠 to 𝑐
of 𝑓 of 𝑥 with respect to 𝑥 plus the limit as 𝑡 tends to positive infinity of
the integral from 𝑐 to 𝑡 of 𝑓 of 𝑥 with respect to 𝑥, where 𝑐 can be any real
number. Now, in our case, 𝑓 of 𝑥 is equal
to 𝑥 multiplied by 𝑒 to the power of negative 𝑥 squared. So we can substitute this into our
formula. Now, we can pick 𝑐 to be any real
number. Let’s choose 𝑐 to be zero. However, you can pick any value of
𝑐 that you like.
Let’s now try to solve our
integrals. And we can do this by trying to
find the antiderivative of 𝑥 multiplied by 𝑒 to the negative 𝑥 squared. We notice that the derivative of 𝑒
to the power of negative 𝑥 squared is equal to negative two 𝑥𝑒 to the power of
negative 𝑥 squared. Now, we can divide both sides of
this equation by the constant term negative two. And we obtain this. On the left-hand side, we have a
constant term multiplying a derivative. Using the derivative rule for
scalar multiplication, we can move the negative one-half inside the derivative. And so, we found our antiderivative
of 𝑥 multiplied by 𝑒 to the power of negative 𝑥 squared. It’s negative one-half 𝑒 to the
power of negative 𝑥 squared.
Using this antiderivative, we’re
able to evaluate the integrals inside our limits. Next, we can substitute in our
upper and lower bounds. We obtained the limit as 𝑠 tends
to negative infinity of negative one-half multiplied by 𝑒 to the power of negative
zero squared minus negative one-half multiplied by 𝑒 to the power of negative 𝑠
squared. Plus the limit as 𝑡 tends to
positive infinity of negative one-half multiplied by 𝑒 to the power of negative 𝑡
squared minus negative one-half multiplied by 𝑒 to the power of negative zero
squared. Since 𝑒 to the power of negative
zero squared is simply 𝑒 to the power of zero, which is just one, the negative
one-half multiplied by 𝑒 to the power of negative zero squared terms is simply
negative one-half. And now, we can sort out the double
negative signs which will give us a positive sign.
Next, we can use one of the
properties of limits, which tells us that the limit of a sum of functions is equal
to the sum of the limits of their functions. And so, we’re able to split our
limits up like this. Here, we have two limits of
constant terms. So these limits are simply equal to
that constant term. The first one being negative
one-half and the second just being one-half. Therefore, when we add these two
terms together, they will cancel out with one another. Now, we just need to worry about
these two remaining limits. Let’s consider the one on the right
first.
As 𝑡 gets larger and larger and
larger, negative 𝑡 squared will be getting larger and larger but in the negative
direction. Therefore, 𝑒 to the power of
negative 𝑡 squared will be getting closer and closer to zero. And so, we can say that the limit
as 𝑡 tends to infinity of negative one-half 𝑒 to the power of negative 𝑡 squared
is equal to zero. Similarly, as 𝑠 gets larger and
larger in the negative direction, 𝑠 squared will be getting larger and larger in
the positive direction. And so, negative 𝑠 squared will be
getting larger in the negative direction. And when this happens, 𝑒 to the
power of negative 𝑠 squared gets closer and closer to zero. Therefore, this limit is also equal
to zero.
Since the negative one-half and the
positive one-half canceled out with one another and these two limits are both equal
to zero, therefore, we can say that the integral from negative infinity to positive
infinity of 𝑥 multiplied by 𝑒 to the power of negative 𝑥 squared with respect to
𝑥 converges to — and is therefore equal to — zero.
