Is there a way to find the antiderivative of $x^k \cos(mx)$?

Question

Is there a way to find the antiderivative of $x^k \cos(mx)$? $k$ and $m$ are just different constants, of course, which may or may not be integers. Integration by parts doesn't lead me anywhere as it just changes the power of $x$ while flipping $\cos$ to $\sin$. I've never been an integral wizard by any stretch, so perhaps there's integration tools I'm not aware of that would apply.

Edit: apologies, did not make clear initially that $x$ can be to a non-integer power. Also, the range of $k$ is between $-1$ and $1$ exclusive. $m$ is positive. Thank you to Very Forgetful Functor for your patience and help. I was hoping to find the form of the antiderative, but was also working through a project where I evaluate this function from $1$ to infinity.

Answer 1

If we had $k\in\mathbb{N}$ then we could find an antiderivative by using integration by parts enough times, but you say that $k\in(-1,1)$ which means that in general there is no way to find this antiderivative in closed form: for instance for $k=\frac{1}{2}$ and $m=1$ we get $$\int x^{\frac{1}{2}}\cos(x) dx$$ so if we substitute $x=t^2$ we find the integral equals $$\int 2t^2 \cos(t^2) dt$$ and two integrations by parts will put us in a situation where we have to calculate $$\int \cos(t^2)dt$$ and it is known that there is no elementary function whose derivative is $\cos(t^2)$.

Answer 2

Hint

$$\int x^k\cos(mx)\,dx=\frac 1{m^{k+1}}\int t^k \cos(t)\,dt$$ $$\int t^k \cos(t)\,dt=\Re\Bigg[\int t^k \,e^{i t}\,dt \Bigg]=\Re\Bigg[(-i)^{k+1} \int u^k \,e^u\,du\Bigg]$$ Now, think about the incomplete gamma function.