I'm studying about Dirac delta function. The Dirac delta distribution is distribution. So it is defined as following: $$\delta[\varphi]=\varphi(0)=\int_{-\infty}^{\infty} \delta(x) \, \varphi(x) \, dx$$
The first line is definition of Dirac delta as distribution, second line means also Dirac delta as distribution but it doesn't mean ordinary integral;it is abuse of notation
And i have seen formula as following: $$\int_{-1}^{4} \delta(x) \, (x^2+1) \, dx$$
The formula above is reasonable? I think 'from -1 to 4' isn't reasonable. Althoug notation in the first picture isn't really mean integral. I think the integral must 'from -infinity to infinity not -1 to 4'
What am I misunderstanding? Or is it related to restriction of distribution? $$\int_{-\infty}^{\infty} (x^2+1)\,\chi_{[-1,4]}(x) \delta(x) \, dx=\int_{-1}^{4} (x^2+1)\, \delta(x) \, dx$$
As Ninad Munshi writes, the integral $$ \int_{-1}^{4} \delta(x) \, (x^2+1) \, dx $$ can be seen as the integral $$ \int_{-\infty}^{\infty} \delta(x) \, (x^2+1) \chi_{[-1,4]}(x) \, dx, $$ where $\chi_{[-1,4]}$ is the indicator function on $[-1,4],$ i.e. it equals $1$ on $[-1,4]$ and $0$ elsewhere.
Now, in general one may not multiply a distribution with a function that is not smooth ($C^\infty$) but when they have disjoint singular support such a product can be defined. Evenmore, for $\delta$ it's enough that the function is continuous at $x=0.$ And so is the case for $(x^2+1) \chi_{[-1,4]}(x)$.
The value of the above integral is of course $0^2+1=1$ since $x\in[-1,4].$