Question

Integrate by parts as we did for (11.14) to obtain (11.15) and (11.16).

Short answer

The integrate by part is $\int \varphi \left(x\right){\delta }^n(x-a)dx=(-1{)}^n\varphi (a{)}^{\left(n\right)}$.

Step-by-step solution

Given information.

The given expression is

$\int \varphi \left(x\right){\delta }^n(x-a)dx$

Definition of Integration By Parts

Integration by partsor partial integration is a process that finds theintegralof aproductoffunctionsin terms of the integral of the product of theirderivativeandantiderivative.

Determine the value of the given expression

The term which is to prove is

$\int \varphi \left(x\right){\delta }^x(x-a)dx={\varphi }^n\left(a\right)$

Consider $\int \varphi \left(x\right){\delta }^n(x-a)dx$integrate by parts

Thus, again the integration by parts multiplied a negative sign, and differentiated the test function once and integrated the delta function one time, thus we expect that using the integration by parts times, we get

$I=(-1{)}^n{\int }_{-\infty }^{\infty }\varphi (x{)}^n\delta (x-a{)}^{(n-n)}dx$

Where,

$\delta (x-a{)}^{\left(0\right)}=\delta (x-a)$

Thus, we have

$I=(-1{)}^n{\int }_{-\infty }^{\infty }\varphi (x{)}^{\left(n\right)}\delta (x-a)dx$

And, we evaluate the integral, thus we have

$I=(-1{)}^n\varphi (a{)}^{\left(n\right)}$

Thus, the integrate by part is $\int \varphi \left(x\right){\delta }^n(x-a)dx=(-1{)}^n\varphi (a{)}^{\left(n\right)}$