# How to find the derivative of improper integral with variable upper limit?

I have the integral from $-\infty$ to $y^2$ of the function $(e^{-|x|})$ and I need to find the derivative of this. That is,

$$\frac{d}{dy} \int_{-\infty}^{y^2} e^{-|x|}\,dx$$

Usually derivative of integral is just the function, but I'm not sure in this case. Should I set up limits or how else should I approach this?

I thought that the derivative of an integral from $-\infty$ to $y^2$ of $(e^{-|x|}dx)$ would be $e^{-|x|}$

• You can define $f(t) = \int_{-\infty}^{t} e^{-|x|}dx$. Then the function you are interested in is $g(y) = f(y^2)$. So you differentiate with the chain rule. Sep 14, 2014 at 18:37
• also note that b/c of the absolute value you need to split the integral. Sep 14, 2014 at 18:39
• @VarunIyer : You could split the integral if you first wanted to solve for the exact integral. But this problem does not require you to do any integration. Sep 14, 2014 at 18:41
• If he would like to know how to solve the integral, its another solution :). Sep 14, 2014 at 18:42
• so lim as a->-infinity of integral from a to 0 e^-|y^2| + integral from 0 to t e^-|y^2| ?
– user176017
Sep 14, 2014 at 18:42

Let $$F(u)=\int_{-\infty}^u e^{-|x|}\,dx.$$ We want $\frac{d}{dy}(F(y^2))$. Let $u=y^2$. By the Chain Rule, we have $$\frac{d}{dy}(F(y^2))=2y\frac{d}{du} F(u).$$

By the Fundamental Theorem of Calculus, we have $\frac{d}{du}F(u)=e^{-|u|}$.

Remark: If the $-\infty$ part makes you uncomfortable, note that $F(y^2)=\int_{-\infty}^0 e^{-|x|}\,dx+\int_0^{y^2} e^{-|x|}\,dx$, and the first integral is just a constant.

This is another possible solution

Much more lengthier, but if you would like to know how to derive:

So the integral is:

$$\int_{-\infty}^{y^2} e^{-|x|} dx = \int_{-\infty}^{0} e^{x} dx + \int_{0}^{y^2} e^{-x} dx$$

In this case, it would be easier to find the integral then perform the derivative:

So for the first we have that:

$$\int_{-\infty}^{0} e^{x} dx = \lim_{b\to-\infty}\int_{b}^{0} e^{x} dx \lim_{b\to-\infty}\left[e^x\right]_{b}^{0} = \lim_{b\to-\infty} 1 - e^b = 1- 0 = 1$$

For the second one:

$$\int_{0}^{y^2} e^{-x} dx = \left[-e^{-x}\right]_{0}^{y^2} = -e^{-y^2} - (-e^{0}) = -e^{-y^2} + 1$$

$$-e^{-y^2} + 1 + 1 = -e^{-y^2} + 2$$
$$\left(-e^{-y^2}\right)' = 2y*e^{-y^2}$$