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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!!!

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  • $\begingroup$ Does $\delta(x)$ denote Dirac function? $\endgroup$ – Lion Mar 4 '14 at 3:33
  • $\begingroup$ Yes, this case,derivative (9) of Dirac function $\endgroup$ – MacArthur Nguyen Mar 4 '14 at 3:36
  • $\begingroup$ This is extraordinarily bad MathJax usage style! You must be using one of those software packages that "help" you with MathJax code. $\endgroup$ – Michael Hardy Mar 4 '14 at 4:23
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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.

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  • $\begingroup$ #Lion Thank you so much, but my problem I'm facing that is: Calculating derivate 9 times of (x^2-4)^9!!! Could you give me more details $\endgroup$ – MacArthur Nguyen Mar 4 '14 at 3:49
  • $\begingroup$ @MacArthurNguyen I add them in my answer. $\endgroup$ – Lion Mar 4 '14 at 4:05
  • $\begingroup$ Thanks Lion very much!!!! ;)) $\endgroup$ – MacArthur Nguyen Mar 4 '14 at 4:49
  • $\begingroup$ @MacArthurNguyen That is my pleasure. $\endgroup$ – Lion Mar 4 '14 at 4:59

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