I have a problem with this integral: $I=\int\limits_{ - 5}^5 ( x^2 - 4 )^9 \delta^{(9)} (x + 2) \, dx $
I did it as this: Use Newton expansion: ${\left( {{x^2} - 4} \right)^9} = {x^{18}} - 9.4{x^{16}} + {36.4^2}{x^{14}} - {84.4^3}.{x^{12}} + {126.4^4}.{x^{10}} - {126.4^5}.{x^8}{\rm{ + 84}}{.4^6}.{x^6} - {36.4^7}.{x^4} + {9.4^8}.{x^2} - {4^9}$ Then, I used formula: $\int \left[ x^n f (x) \right] \delta ^{\left( n \right)}(x) \,dx = (-1)^n n! \int f(x)\delta (x) \, dx $
And the final result: $I=0$ (Wrong or Right??)
But my teacher said I should do it agian??
Thanks my friends!!!
The Dirac function have the following property: \begin{equation} \int_a^bf(x)\delta^{(n)}(x-c)dx=(-1)^nf^{(n)}(x)|_{x=c}\quad a<c<b \end{equation} So your problem can be solved by: \begin{equation} \int_{-5}^5(x^2-4)^9\delta^{(9)}(x+2)dx=(-1)^9[(x^2-4)^9]^{(9)}|_{x=-2}=95126814720 \end{equation}
About $[(x^2-4)^9]^{(9)}$ We can use binomial expansion theorem: \begin{equation} (x^2-4)^9=\sum_{n=0}^9C_9^nx^{2n}(-4)^{9-n} \end{equation} Then the derivative is \begin{equation} [(x^2-4)^9]^{(9)}=(\sum_{n=0}^9C_9^nx^{2n}(-4)^{9-n})^{(9)}=\sum_{n=5}^9C_9^n(x^{2n})^{(9)}(-4)^{9-n}\\ =\sum_{n=5}^9C_9^n(A_{2n}^9x^{2n-9})(-4)^{9-n} \end{equation} where $C_n^m$ is number of permutation and $A_n^m$ is number of arrangement.