Solving integral equation with Laplace's Transform. I'm trying to prove the following
$$\int\limits_0^\infty  {\frac{{\cos tu}}{{{u^2} + 1}}\log udu}  =  - \frac{\pi }{2}\int\limits_0^\infty  {\frac{{\sin tu}}{{{u^2} + 1}}du} $$
The original problem suggested the use of the Laplace Transform, thus I have
$$\int\limits_0^\infty  {\frac{s}{{{u^2} + {s^2}}}\frac{{\log u}}{{{u^2} + 1}}du = }  - \frac{\pi }{2}\int\limits_0^\infty  {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$
ADD The RHS transform can be evaluated as follows:
$$- \frac{\pi }{2}\int\limits_0^\infty  {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$
$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} $$
Now by partial fractions you can separate and get:
$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}}  =  - \frac{\pi }{4}\frac{1}{{{s^2} - 1}}\left[ {\log \left( {\frac{{m + 1}}{{m + {s^2}}}} \right)} \right]_0^\infty $$
$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}}  = -\frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
For the second one I make a similar manipulation:
$$\frac{1}{{{s^2} + {u^2}}}\frac{1}{{{u^2} + 1}} = \frac{1}{{{s^2} - 1}}\left( {\frac{1}{{{u^2} + 1}} - \frac{1}{{{u^2} + {s^2}}}} \right)$$
$$\frac{s}{{{s^2} - 1}}\int\limits_0^\infty  {\left( {\frac{{\log u}}{{{u^2} + 1}} - \frac{{\log u}}{{{u^2} + {s^2}}}} \right)du} $$
Substitution to show integral over $\mathbb{R}$ of an odd function is zero so,
$$\frac{s}{{{s^2} - 1}}\left( {\int\limits_{ - \infty }^\infty  {\frac{{x{e^x}}}{{{e^{2x}} + 1}}dx}  - \int\limits_0^\infty  {\frac{{\log u}}{{{u^2} + {s^2}}}} }du \right)$$
$$ - \frac{s}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{{\log u}}{{{u^2} + {s^2}}}}du $$
And again a suitable $u = m s$ produces
$$ - \frac{1}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{{\log m + \log s}}{{{m^2} + 1}}} dm =  - \frac{{\log s}}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{1}{{{m^2} + 1}}} dm =  - \frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
 A: $$- \frac{\pi }{2}\int\limits_0^\infty  {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$$
$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}} $$
Now by partial fractions you can separate and get:
$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}}  =  - \frac{\pi }{4}\frac{1}{{{s^2} - 1}}\left[ {\log \left( {\frac{{m + 1}}{{m + {s^2}}}} \right)} \right]_0^\infty $$
$$ - \frac{\pi }{4}\int\limits_0^\infty  {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}}  = -\frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
For the second one I make a similar manipulation:
$$\frac{1}{{{s^2} + {u^2}}}\frac{1}{{{u^2} + 1}} = \frac{1}{{{s^2} - 1}}\left( {\frac{1}{{{u^2} + 1}} - \frac{1}{{{u^2} + {s^2}}}} \right)$$
$$\frac{s}{{{s^2} - 1}}\int\limits_0^\infty  {\left( {\frac{{\log u}}{{{u^2} + 1}} - \frac{{\log u}}{{{u^2} + {s^2}}}} \right)du} $$
Substitution to show integral over $\mathbb{R}$ of an odd function is zero so,
$$\frac{s}{{{s^2} - 1}}\left( {\int\limits_{ - \infty }^\infty  {\frac{{x{e^x}}}{{{e^{2x}} + 1}}dx}  - \int\limits_0^\infty  {\frac{{\log u}}{{{u^2} + {s^2}}}} }du \right)$$
$$ - \frac{s}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{{\log u}}{{{u^2} + {s^2}}}}du $$
And again a suitable $u = m s$ produces
$$ - \frac{1}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{{\log m + \log s}}{{{m^2} + 1}}} dm =  - \frac{{\log s}}{{{s^2} - 1}}\int\limits_0^\infty  {\frac{1}{{{m^2} + 1}}} dm =  - \frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$
