Help on integral regarding analytical Fourier transform

Question

To explain my problem I start with two functions to be sine transformed. This question is a problem of current research in the field of electrolyte transport theory. The first Function is given by:

$$Y_1(r)=\frac{1}{r\cdot\biggl[\text{e}^{+A\cdot r}+B\cdot\text{e}^{-C\cdot r}\biggr]}$$

and

$$Y_2(r)=\frac{1}{r^2\cdot\biggl[\text{e}^{+A\cdot r}+B\cdot\text{e}^{-C\cdot r}\biggr]}$$

with constants $A,B,C\in\mathcal{R}^+$ beeing positive real numbers without 0. The symbol $r\in\mathcal{R}^+_0$ denotes the variable in coordinate space. Note that the important condition $C>A$ holds! The sine transform of the two functions are given by:

$$Z_i(k) = \int_{0}^{\infty}Y_i(r)\cdot\text{sin}(k\cdot r)~\text{d}r\quad\forall\quad i\in[1,2]$$

with $k\in\mathcal{R}^+_0$ beeing the variable in Fourier space. The two integrals should exist unter the condition $C>A$.

But I'm having trouble solving the integrals. Unfortunately, I haven't been able to find a solution using Maple either. The integral tables available to me didn't help me either. Can anyone solve these two integrals or know of literature sources with corresponding solutions? Can someone perhaps solve these two integrals with another CAS like Mathematica? If a solution is possible, I would be very happy if the solution could be roughly outlined. Thank you in advance.

Answer 1

Consider the whole thing a function of $A$. Differentiating this function boils the first integral down to $$ \frac A{2i}\int_0^\infty (e^{ix}-e^{-ix})\frac{e^{Ax}}{e^{Ax}+Be^{-Cx}}\ dx =\frac A{2i}\int_0^\infty (e^{ix}-e^{-ix})\frac{1}{1+Be^{(-A-C)x}}\ dx $$ Now for $B<1$ expand the second term in the integral as a geometric series and interchange the sum with the integral. The upcoming Integrals are elementary and after differentiating you get a geometric series again which reduces the whole thing to a rational function in $e^{(A+C)x}$. Hence the orgiginal function is an double integral of a very simple rational function which can be easily evaluated.

Answer 2

The integral $Y_2$ is divergent (it has a non-integrable pole at $r=0$).

Concerning $Y_1$, Mathematica gives a complicated closed-form answer for $B=1$, in terms of the generalised Stieltjes constant $\gamma_n(a)$, the Euler gamma constant $\gamma$, the Gamma function $\Gamma(s)$, and the zeta function $\zeta(s)$: _{$$\int_0^\infty\frac{\sin k r}{r (e^{Ar}+e^{-Cr})}\,dr$$ $$=\frac{1}{2(A+C)}\Re\biggl[(k+i A) \gamma _1\left(\frac{A-i k}{2 (A+C)}\right)-(k+i A) \gamma _1\left(X\right)+2 i (A+C) \biggl((\ln (A+C)+\gamma +\ln 2) \left[\zeta \left(0,X\right)-\zeta \left(0,Y\right)\right]+\ln \Gamma \left(X\right)+\ln \Gamma \left(Y\right)\biggr)\biggr],$$ $$X=\frac{3 A+2 C-i k}{2 (A+C)},\;\;Y=\frac{2 A+C-i k}{2 (A+C)}.$$}