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:

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

Thanks for your answers

  • 4
    $\begingroup$ Hm, I don't understand how you have a function of $x$ as the result. $\endgroup$ – Simply Beautiful Art Dec 30 '16 at 23:58
  • 4
    $\begingroup$ 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 $g(x) = e^{-x}$. $\endgroup$ – Winther Dec 31 '16 at 0:01
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    $\begingroup$ The answer should be $1+\tfrac{1}{2}e^{-1}+\tfrac{1}{2}e$. $\endgroup$ – JimmyK4542 Dec 31 '16 at 0:01
  • $\begingroup$ @JimmyK4542 yeah that's why I think but I'm new in this so I don't know if I miss something $\endgroup$ – rpestana94 Dec 31 '16 at 0:08

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: $$ \int_{-\infty}^\infty f(x)\delta(x-a)\,\mathrm{d}x = f(a). $$

Breaking up the integral with linearity, we get \begin{align*} \int_{-\infty}^{\infty} \Big[\delta(x)+ &\frac{\delta(x-1)}{2} + \frac{\delta(x+1)}{2}\Big] e^{-x}\,\mathrm{d}x \\ &=\int_{-\infty}^{\infty} \delta(x)e^{-x}\,\mathrm{d}x+ \int_{-\infty}^\infty\frac{\delta(x-1)}{2}e^{-x}\,\mathrm{d}x + \int_{-\infty}^\infty\frac{\delta(x+1)}{2}e^{-x}\,\mathrm{d}x. \end{align*} Applying the sifting property gives us the answer others have mentioned.


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