Finding the integral of $\int_{-\infty}^{\infty}e^{-|4x|}$.

Question

So I am trying to find the integral of $\int_{-\infty}^{\infty}e^{-|4x|}$. I know the integral converges, and I know the answer as well, but I am confused on how to get the correct answer. My problem I suspect is coming from correctly integrating with the absolute value in the exponent. So I'll post my steps and someone point my mistake, thanks. $$\int_{-\infty}^{\infty}e^{-|4x|}\rightarrow \lim_{a\rightarrow -\infty}\int_{a}^{0} e^{-|4x|} + \lim_{b\rightarrow \infty}\int_{0}^{b} e^{-|4x|} $$

I integrate by substion here, which is where I feel I am making my mistake, but I do not know if I am appropriately substituting in regards to the absolute value. $$u=-|4x|\hspace{5pt}du=-4dx?$$ $$\lim_{a\rightarrow -\infty} -\frac{1}{4}e^{-|4x|}|_{a}^{0}\hspace{10pt}+\hspace{10pt}\lim_{b\rightarrow -\infty} -\frac{1}{4}e^{-|4x|}|_{0}^{b}$$

$$\lim_{a\rightarrow -\infty}\left( -\frac{1}{4}+\frac{1}{4}e^{-|4\infty|} \right)+ \lim_{b\rightarrow \infty} \left( -\frac{1}{4}e^{-|4(-\infty|)} +\frac{1}{4} \right) $$

$$-\frac{1}{4}+0-0+\frac{1}{4}$$ So I got $0$ but the answer is $\frac{1}{2}$ so I am thinking the absolute value is supposed to go onto the whole integral. Ex- $\int e^{-|4x|}=\left| -\frac{1}{4}e^{-|4x|} \right|$ or just $\left| \frac{1}{4}e^{-|4x|} \right|$. Is my assumption correct, or what is the correct way of doing this integral, thanks in advance!

Answer 1

$$\int_{-\infty}^{\infty}e^{-|4x|}dx=\int_{-\infty}^{0}e^{-|4x|}dx+\int_{0}^{\infty}e^{-|4x|}dx$$ $$=\int_{-\infty}^{0}e^{4x}dx+\int_{0}^{\infty}e^{-4x}dx=2\int_{0}^{\infty}e^{-4x}dx=1/2$$

Answer 2

Your derivative is wrong. While it would be a worthwhile exercise to work carefully how you were computing the derivative and figure out where you went wrong, an easier approach for just getting the answer is to realize you can simplify away all of the appearances of the absolute value.

Answer 3

Since the function $e^{-4|x|}$ is even we have $$\int_{-\infty}^{\infty}e^{-4|x|}\, dx=2\int_{0}^{\infty}e^{-4x}\, dx$$ On the other hand $$2\int_{0}^{\infty}e^{-4x}\, dx=2\lim\limits_ {a\to \infty }\int_{0}^{a}e^{-4x}\, dx=\frac{1}{2}$$