for the following integral: $$\int_{0}^{\infty} \frac{\sin\!\left(k\, x\right)\, \left(\mathrm{e}^{- a\, x} - \mathrm{e}^{- b\, x}\right)}{x^2} \,dx$$ The following analytic solution can be obtained: $$\frac{k\, \mathrm{log}\!\left(b^2 + k^2\right)}{2} - \frac{k\, \mathrm{log}\!\left(a^2 + k^2\right)}{2} - a\, \mathop{\mathrm{arccot}}\nolimits\!\left(\frac{a}{k}\right) + b\, \mathop{\mathrm{arccot}}\nolimits\!\left(\frac{b}{k}\right) $$
I was wondering how we can obtain the solution for the following integral: $$\int_{0}^{\infty} \frac{\sin\!\left(k\, x\right)\, \left(C\, \mathrm{e}^{- a\, x} - D\, \mathrm{e}^{- b\, x}\right)}{x^2} \,d x $$ Where C and D are constants. My question can be extended for higher x powers. for example for the following integral: $$\int_{0}^{\infty} \frac{\sin\!\left(k\, x\right)\, \left(\mathrm{e}^{- a\, x} - \mathrm{e}^{- b\, x}\right)}{x^4} \,d x $$ The analytical solution is: $$\frac{b^3\, \mathop{\mathrm{arccot}}\nolimits\!\left(\frac{b}{k}\right)}{6} - \frac{a^3\, \mathop{\mathrm{arccot}}\nolimits\!\left(\frac{a}{k}\right)}{6} + \frac{k^3\, \mathrm{log}\!\left(a^2 + k^2\right)}{12} - \frac{k^3\, \mathrm{log}\!\left(b^2 + k^2\right)}{12} + \frac{5\, a^2\, k}{12} - \frac{5\, b^2\, k}{12} - \frac{a\, k^2\, \arctan\!\left(\frac{a}{k}\right)}{2} + \frac{b\, k^2\, \arctan\!\left(\frac{b}{k}\right)}{2} - \frac{a^2\, k\, \mathrm{log}\!\left(a^2 + k^2\right)}{4} + \frac{b^2\, k\, \mathrm{log}\!\left(b^2 + k^2\right)}{4} $$
How can we solve the integral, if we multiply each term by different constants(C and D), as we did for the first case?
Thanks,