Evaluate $f_n(\alpha,\beta)=\int_0^{\infty}\mathrm{e}^{-x^n}\sin(\alpha x)\cos(\beta x)\,dx$ I was just looking at the following function but couldn’t understand how to form a way of integrating this and how does it depend upon $n, \alpha$ and $\beta$ can anyone please help
$$f_n(\alpha,\beta)=\int_0^{\infty}\mathrm{e}^{-x^n}\sin(\alpha x)\cos(\beta x)\,dx$$ 
 A: Well, we have the following integral:
$$\mathcal{I}_\text{n}\left(\alpha,\beta\right):=\int_0^\infty\exp\left(-x^\text{n}\right)\sin\left(\alpha x\right)\cos\left(\beta x\right)\space\text{d}x\tag1$$
Using the definition of the Exponential function:
$$\exp(x)=\sum_{\text{k}\ge0}\frac{x^\text{k}}{\text{k}!}\tag2$$
So, we can write:
$$\mathcal{I}_\text{n}\left(\alpha,\beta\right)=\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}}{\text{k}!}\int_0^\infty x^\text{kn}\sin\left(\alpha x\right)\cos\left(\beta x\right)\space\text{d}x\tag3$$
Now, we also know that:
$$\sin\left(\alpha x\right)\cos\left(\beta x\right)=\frac{\sin\left(\left(\alpha-\beta\right)x\right)+\sin\left(\left(\alpha+\beta\right)x\right)}{2}\tag4$$
So:
$$\mathcal{I}_\text{n}\left(\alpha,\beta\right)=\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}}{2\left(\text{k}!\right)}\left\{\underbrace{\int_0^\infty x^\text{kn}\sin\left(\left(\alpha-\beta\right)x\right)\space\text{d}x}_{\text{I}_1}+\underbrace{\int_0^\infty x^\text{kn}\sin\left(\left(\alpha+\beta\right)x\right)\space\text{d}x}_{\text{I}_2}\right\}\tag5$$
Now, we can use the 'evaluating integrals over the positive real axis' property of the Laplace transform in order to write:


*

*$$\text{I}_1=\int_0^\infty\mathcal{L}_x\left[\sin\left(\left(\alpha-\beta\right)x\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[x^\text{kn}\right]_{\left(\text{s}\right)}\space\text{ds}\tag6$$

*$$\text{I}_2=\int_0^\infty\mathcal{L}_x\left[\sin\left(\left(\alpha+\beta\right)x\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[x^\text{kn}\right]_{\left(\text{s}\right)}\space\text{ds}\tag7$$
And using the table of selected Laplace transforms, we have:


*

*$$\mathcal{L}_x\left[\sin\left(\left(\alpha-\beta\right)x\right)\right]_{\left(\text{s}\right)}=\frac{\alpha-\beta}{\left(\alpha-\beta\right)^2+\text{s}^2}\tag8$$

*$$\mathcal{L}_x\left[\sin\left(\left(\alpha+\beta\right)x\right)\right]_{\left(\text{s}\right)}=\frac{\alpha+\beta}{\left(\alpha+\beta\right)^2+\text{s}^2}\tag9$$

*$$\mathcal{L}_x^{-1}\left[x^\text{kn}\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}^{1+\text{kn}}}\cdot\frac{1}{\Gamma\left(-\text{kn}\right)}\tag{10}$$

In order to finish you can use this.

