How to compute $\int_{1}^{\infty} dx \sin(\beta x)e^{-\alpha x}x^n $or express it in terms of special functions? Compute $$\int_{1}^{\infty} dx \sin(\beta x) e^{-\alpha x}x^n,$$ where $\beta ,\alpha >0$, $n$ is an integer (positive or negative).
I am not quite familiar with special functions, any help would be appreciated!
 A: Looking at the more general integral
$$\int_{\gamma}^{\infty} dx \sin(\beta x) e^{-\alpha x}x^n,$$
we can rewrite this using differentiation under the integral sign:
$$(-1)^n\frac{\mathrm{d}^n}{\mathrm{d}\alpha^n}\int_{\gamma}^{\infty} \mathrm{d}x \sin(\beta x) e^{-\alpha x}.$$
To compute the integral we use $\sin(x)=\operatorname{Im}(e^{\mathrm{i}x})$ and arrive at
$$(-1)^n\frac{\mathrm{d}^n}{\mathrm{d}\alpha^n}\operatorname{Im}\int_{\gamma}^{\infty} \mathrm{d}x\,e^{-(\alpha-\mathrm{i}\beta) x}$$
which is straight forward and yields
$$(-1)^n\frac{\mathrm{d}^n}{\mathrm{d}\alpha^n}\left(\frac{\mathrm{e}^{-\alpha\gamma}\cdot\left(\alpha\sin(\beta\gamma)+b\cos(\beta\gamma)\right)}{\alpha^2+\beta^2}\right).$$
At this point though I don't know how to express the derivative above in a closed form, but maybe this helps.
Edit: It is possible to simplify the expression above a little bit further: first of all we can use the harmonic addition theorem to write the sum of $\cos$ and $\sin$ as one $\cos$:
$$(-1)^n\frac{\mathrm{d}^n}{\mathrm{d}\alpha^n}\left[\frac{\mathrm{e}^{-\alpha\gamma}}{\sqrt{\alpha^2+\beta^2}}\cdot\cos\left(\beta\gamma-\arctan\left(\frac{\alpha}{\beta}\right)\right)\right]$$
and then use the shift theorem to "pull out" the exponential:
$$(-1)^n\mathrm{e}^{-\alpha\gamma}\left(\frac{\mathrm{d}}{\mathrm{d}\alpha}-\gamma\right)^n\left[\frac{1}{\sqrt{\alpha^2+\beta^2}}\cdot\cos\left(\beta\gamma-\arctan\left(\frac{\alpha}{\beta}\right)\right)\right].$$
Again, this is a long way off a general expression, but it should make things easier.
A: We shall exploit a Laplace transform integral identity
$$I=\int_{0}^{\infty} \sin \left(\beta (x+1)\right) e^{-\alpha (x+1)} (x+1)^n \, dx = \int_{0}^{\infty} \mathcal{L}\left[\sin \left(\beta (x+1)\right) e^{-\alpha (x+1)}\right](s) \cdot \mathcal{L}^{-1}\left[(x+1)^n\right] (s) \, ds$$
Since
$$\mathcal{L}^{-1} \left[(x+1)^n\right]= \frac{e^{-s} s^{-1-n}}{\Gamma (-n)}$$
and
$$\mathcal{L} \left[ \sin (\beta (x+1)) e^{-\alpha (x+1)}\right] = \frac{e^{-\alpha} \left(\beta \cos (\beta) +(\alpha +s) \sin (\beta)\right)}{\beta^2+(\alpha+s)^2}$$
So
$$I = \int_{0}^{\infty}\frac{e^{-\alpha-s} s^{-1-n} (\beta \cos (\beta) +(\alpha + s) \sin (\beta)}{\left(\beta^2 + (\alpha+s)^2\right) \Gamma (-n)} \, ds$$
Now Mathematica gives the following Mellin transform- I presume one can prove it with Ramanujan’s master theorem:
$$\int_{0}^{\infty} s^{t-1} \left( \frac{e^{-\alpha-s} (\beta \cos (\beta) +(\alpha + s) \sin (\beta)}{\left(\beta^2 + (\alpha+s)^2\right) \Gamma (-n)} \right) \, ds = \frac{e^{-\alpha}\Gamma(t)\left(i t e^\alpha (\alpha+i\beta)E_{t+1}(\alpha-i\beta)-t e^{\alpha}(\beta+i\alpha)E_{t+1}(\alpha+i\beta)+2(\alpha\sin(\beta)+\beta\cos(\beta))\right)}{2\left(\alpha^2+\beta^2\right)\Gamma(-n)}$$
Where $E_n$ is the generalised exponential integral function.
Now take $t \mapsto -n$ and we determine:
$$\boxed{I = \frac{e^{-\alpha}}{2\left(\alpha^2+\beta^2\right)}\left(2\alpha\sin(\beta)+2\beta\cos(\beta)+n e^{\alpha}(\beta-i\alpha)E_{1-n}(\alpha-i\beta)+n e^{\alpha}(\beta+i\alpha)E_{1-n}(\alpha+i\beta)\right)}$$
A: We are going to evaluate the improper integral
$$ I=\int_1^\infty x^ne^{-\alpha x}\sin\beta x\,\mathrm dx$$
I will assume that $n\in\mathbb Z_{\ge0}$.
We substitute $x\mapsto x+1$, use the binomial theorem and then the angle sum formula for sine.
$$\begin{align}I &=\int_0^\infty(x+1)^ne^{-\alpha(x+1)}\sin\beta(x+1)\,\mathrm dx\\&=e^{-\alpha}\sum_{k=0}^n\binom nk\int_0^\infty x^ke^{-\alpha x}[\sin\beta x\cos\beta+\cos\beta x\sin\beta]\,\mathrm dx \\ I&= e^{-\alpha}\sum_{k=0}^n(-1)^k\binom nk\Big[\cos\beta \frac{\partial^k}{\partial\alpha^k} \frac\beta{\alpha^2+\beta^2}+\sin\beta \frac{\partial^k}{\partial\alpha^k} \frac\alpha{\alpha^2+\beta^2}\Big]\end{align}$$
Now, using the partial fraction decompositions
$$\begin{align} \frac\beta{\alpha^2+ \beta^2} &=\frac1{2i}\Big[\frac1{\alpha-i\beta}-\frac1{\alpha+i\beta}\Big]\\ \frac\alpha{\alpha^2+\beta^2}&= \frac12\Big[\frac1{\alpha-i\beta}+\frac1{\alpha+i\beta}\Big] \end{align}$$
we get
$$\begin{align} \frac{\partial^k}{\partial\alpha^k}\frac\alpha{\alpha^2+\beta^2}&=\frac{ (-1)^kk!}2\bigg[\frac1{(\alpha-i\beta)^{k+1}}+\frac1{(\alpha+i\beta)^{k+1}} \bigg]\\ &= \frac{(-1)^kk!}{(\alpha^2+\beta^2)^{\frac{k+1}2}}\bigg[\frac{\exp\big(i(k+1)\arctan\frac\beta\alpha\big)+\exp\big(-i(k+1)\arctan\frac\beta\alpha\big)}2\bigg] \\ \frac{\partial^k}{\partial\alpha^k}\frac\alpha{\alpha^2+\beta^2}&= \frac{(-1)^kk!}{(\alpha^2+\beta^2)^{\frac{k+1}2}}\cos(k+1)\arctan\frac\beta\alpha\end{align}$$
and similarly
$$\frac{\partial^k}{\partial\alpha^k}\frac{\beta}{\alpha^2+\beta^2}= \frac{(-1)^kk!}{(\alpha^2+\beta^2)^{\frac{k+1}2}}\sin(k+1)\arctan\frac\beta\alpha$$
Using these results, we get our final closed form
$$\boxed{\boxed{ \int_1^\infty x^ne^{-\alpha x}\sin\beta x\,\mathrm dx =e^{-\alpha} \sum_{k=0}^n\frac{n!}{(n-k)!} \frac1{(\alpha^2+\beta^2)^{\frac{k+1}2}}\sin\Big[(k+1)\arctan\frac\beta\alpha+\beta\Big]}}$$
