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

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!

• When you have such cases, split the problem in two pieces – Claude Leibovici Sep 14 '14 at 7:14

$$\int_{-\infty}^{\infty}e^{-|4x|}dx=\int_{-\infty}^{0}e^{-|4x|}dx+\int_{0}^{\infty}e^{-|4x|}dx$$ $$=\int_{-\infty}^{0}e^{4x}dx+\int_{0}^{\infty}e^{-4x}dx=2\int_{0}^{\infty}e^{-4x}dx=1/2$$
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 $$2\int_{0}^{\infty}e^{-4x}\, dx=2\lim\limits_ {a\to \infty }\int_{0}^{a}e^{-4x}\, dx=\frac{1}{2}$$