$$\int_{-\infty}^{\infty}\frac{x\sin(ax)}{(x^{2}-b^{2})(x^{2}+c^{2})}dx=\pi(\frac{\cos(ab)-e^{-ac}}{b^{2}+c^{2}})$$
Where $a,b,c \in \mathbb{R^{+}}$ I was wondering if anyone can confirm If the above is correct, I don't need a solution just verification, note that the above is Gradshteyn and Ryzhik equation 3.728.9 but there solution is incorrect in the book of table of integrals, series and products (7th edition ) just a comment would be sufficient
Notice that via residues theorem we can rewrite it as
$$\int_{-\infty}^{+\infty} \frac{z \Im e^{iaz}}{(z^2-b^2)(z+ic)(z-ic)}\ dz$$
A simple pole is present at $x = \pm ic$ whilst only one lies in the upper half plane whence
$$2\pi i \Im \left(\frac{z e^{iaz}}{(z^2-b^2)(z+ic)}\right)\bigg|_{z = ic}$$
Which gives
$$\color{red}{-\pi \frac{e^{-ac}}{b^2+c^2}}$$
NOT THE END
This is a "special" integral. It's not only made of the usual poles on the imaginary axis, but real poles along the real axis are present. Indeed we also have $x = \pm b$. Real.
In this case, residue theorem guarantees an additional term in the solution which is:
$$\pi i \text{Res }\ \left(f(x)e^{iax}\right)\bigg|_{x_k}$$
Where this time the poles are real.
By substituting $x_k = \pm b$ we get
$$\pi i \left(\frac{b e^{iab}}{(b^2+c^2)(2b)} + \frac{(-b)e^{-iab}}{(b^2+c^2)(-2b)}\right)$$
Using elementary trigonometry, arranging, we get eventually
$$i\pi \frac{e^{iab} + e^{-iab}}{2(b^2+c^2)}$$
Hence
$$i\pi \frac{\cos(ab)}{b^2+c^2}$$
Being it part of the Imaginary request part, we get the additional term
$$\pi \frac{\cos(ab)}{b^2+c^2}$$
Which summed to the previous term gives really
$$\color{blue}{\pi\left(\frac{\cos(ab) - e^{-ac}}{b^2+c^2}\right)}$$
THIS MEANS THE ABOVE SOLUTION IS CORRECT