Convergent integral of divergent exponential function $$$$
The following integral is a divergent integral:
$$\int_{0}^{\infty} \frac{\mathrm{e}^{- a\, x}\, \sin\!\left(x\right)}{x^5} \,d x $$
However, the following solution is provided (though divergent solution) in the book of "Theory of elasticity of microheterogeneous media", by Shermergor (in Russian):
$$\frac{25\, a}{72} + \frac{\mathrm{\delta}\, \left(3\, a^2 - 1\right)}{6} - \frac{a\, {\mathrm{\delta}}^2}{2} - \frac{25\, a^3}{72} + \frac{\arctan\!\left(\frac{1}{a}\right)\, \left(a^4 - 6\, a^2 + 1\right)}{24} + \frac{{\mathrm{\delta}}^3}{3} + \frac{a\, \mathrm{log}\!\left(a^2 + 1\right)\, \left(a^2 - 1\right)}{12}$$
Where $\mathrm{\delta}$ is an infinite constant.  Can anybody help me to understand how this solution is obtained?
Is there any logical hierarchy to obtain analogous solutions for the general case of:
$$\int_{0}^{\infty} \frac{\mathrm{e}^{- a\, x}\, \sin\!\left(x\right)}{x^n} \,d x $$
Thank you so much for the time,
 A: Well, we are looking at:
$$\mathcal{I}_{\space\text{n}}\left(\alpha\right):=\int_0^\infty\frac{\exp\left(-\alpha\cdot x\right)\cdot\sin\left(x\right)}{x^\text{n}}\space\text{d}x\tag1$$
Using the evaluating integrals over the positive real axis property of the Laplace transform we can write:
$$\mathcal{I}_{\space\text{n}}\left(\alpha\right)=\int_0^\infty\mathscr{L}_x\left[\exp\left(-\alpha\cdot x\right)\cdot\sin\left(x\right)\right]_{\left(\text{s}\right)}\cdot\mathscr{L}_x^{-1}\left[\frac{1}{x^\text{n}}\right]_{\left(\text{s}\right)}\space\text{d}\text{s}\tag2$$
Using the table of selected Laplace transforms, we can write:


*

*$$\mathscr{L}_x\left[\exp\left(-\alpha\cdot x\right)\cdot\sin\left(x\right)\right]_{\left(\text{s}\right)}=\frac{1}{1+\left(\alpha+\text{s}\right)^2}\tag3$$

*$$\mathscr{L}_x^{-1}\left[\frac{1}{x^\text{n}}\right]_{\left(\text{s}\right)}=\frac{\text{s}^{\text{n}-1}}{\Gamma\left(\text{n}\right)}\tag4$$


So, we get:
$$\mathcal{I}_{\space\text{n}}\left(\alpha\right)=\int_0^\infty\frac{1}{1+\left(\alpha+\text{s}\right)^2}\cdot\frac{\text{s}^{\text{n}-1}}{\Gamma\left(\text{n}\right)}\space\text{d}\text{s}=\frac{1}{\Gamma\left(\text{n}\right)}\cdot\int_0^\infty\frac{\text{s}^{\text{n}-1}}{1+\left(\alpha+\text{s}\right)^2}\space\text{d}\text{s}\tag5$$
