Applying differentiation with respect to a parameter, evaluate the following integral:
$$\int_{0}^{\infty} e^{-\alpha x} \frac{\sin(\beta x)}{x} dx$$
I am struggling to solve this, furthermore, I do not understand their suggested approach.
if we create the function $F(\alpha, \beta) = \int_{0}^{\infty} e^{-\alpha x} \frac{\sin(\beta x)}{x} dx$. If we find
$$\frac{\partial}{\partial \alpha}F(\alpha, \beta) = f(\alpha), \frac{\partial}{\partial \beta}F(\alpha, \beta) = f(\beta)$$
So then $F(\alpha, \beta) = \int f(\alpha) d\alpha + g_1(\beta)$ and $F(\alpha, \beta) = \int f(\beta) d\beta + g_1(\alpha)$ ?? I am not sure how we will be able to find the left over constant term here ....
Edit:
The last answer provided here: Evaluating $\int_0^\infty e^{-ax}\frac{\sin{(bx)}}{x}dx $
Provides a bit a starting point. We calculate the following:
$$\frac{\partial}{\partial a} \int_{0}^{\infty} e^{-ax} \frac{ \sin(bx)}{x}dx = -\frac{b}{a^2+b^2}$$
$$\frac{\partial}{\partial b} \int_{0}^{\infty} e^{-ax} \frac{ \sin(bx)}{x}dx = \frac{a}{a^2+b^2}$$
Now integrating with respect to the corresponding variables gives me $\frac{-1}{b}\arctan(\frac{a}{b}) +C_1$ and $\frac{1}{a}\arctan(\frac{b}{a})+C_2$. How do I go about finding the functions $C_1$ and $C_2$?
