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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}}$

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    @Lord_Farin Do you have any other solution?2013-05-24

1 Answers 1

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$- \frac{\pi }{2}\int\limits_0^\infty {\frac{u}{{{s^2} + {u^2}}}\frac{{du}}{{{u^2} + 1}}}$

$ - \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}}$