indefinite integral problem $\int x^n \cos(mx) \text{d}x$

Question

I was going through my old calculus book and found a integral problem, i generalised the problem and tried to solve it.I am not sure if the solution is correct and also what would happen if n tends to infinity. Is there any other method to solve this problem? please refer the below link https://1drv.ms/f/s!AuU4aSVY7XTLcPp2rr8x0Cf6ZLc

Answer 1

Is there any other method to solve this problem?

Yes.

Hint. One may start with $$ \int e^{(a+im)x}dx=\frac{e^{(a+im)x}}{a+im},\qquad \qquad (a,m)\in \mathbb{R}^2\backslash\left\{(0,0)\right\}, $$ then one may differentiate $n$ times with respect to $a$, using the general Leibniz rule for a product and considering $a=0$ in the real part of each side.

Answer 2

It s more convenient to study the exponential form (of which f is the real part)

$$g(\text{n},\text{m},x)\text{=}\int x^n \exp (i m x) \, dx$$

and observe that the factor $x^n$ can be generated by n-fold differentiation of the exponential with respect to m.

Exchanging integration and differentiation, and carrying out the integration over $x$ for $n = 0$ gives for $g$

$$\frac{\frac{\partial ^n}{\partial m^n}\left(-\frac{i e^{i m x}}{m}\right)}{(i x)^n}$$

Carrying out the differentiation according to the Leibniz rule leads to the expression

$$g=-i e^{i m x} \sum _{k=0}^n \frac{(-1)^k k! \binom{n}{k} (i x)^{n-k}}{m^{k+1}}$$

$g$ can also be expressed in compact form using the incomplete $\Gamma$ function

$$g = -\frac{i (i x)^n (-i m x)^{-n} \Gamma (n+1,-i m x)}{m}$$

$f$ is then given by the real part.