# Integration of dirac, why this is the result?

I don't understand why this example:

$$\int_{-\infty}^{\infty} \left(\ \delta(x)+ \frac{\delta(x-1)}{2} + \frac{\delta(x+1)}{2}\right)\ e^{-x} \ dx$$

Gives the following anwser:

$$1 \ + \ \frac{e^{-x}}{2} \ + \frac{e^{x}}{2}$$

• Hm, I don't understand how you have a function of $x$ as the result. – Simply Beautiful Art Dec 30 '16 at 23:58
• Note that $\int_{-\infty}^\infty\delta(x-1)g(x){\rm d}x = g(1)$ and $\int_{-\infty}^\infty\delta(x+1)g(x){\rm d}x = g(-1)$ while you seem to have plugged in $g(x)$ and $g(-x)$ respectively. The result should be $1 + \frac{e^{-1}}{2} + \frac{e^1}{2} = g(0) + \frac{g(1)}{2} + \frac{g(-1)}{2}$ where $g(x) = e^{-x}$. – Winther Dec 31 '16 at 0:01
• The answer should be $1+\tfrac{1}{2}e^{-1}+\tfrac{1}{2}e$. – JimmyK4542 Dec 31 '16 at 0:01
As others pointed out in the comments, this answer is certainly not correct because we expect a real number as the evaluation of the definite integral, not a function. The key is to apply the "sifting" property of the delta function: $$\int_{-\infty}^\infty f(x)\delta(x-a)\,\mathrm{d}x = f(a).$$