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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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    The integral you have doesn't converge.2012-02-01
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    Something that might salvage your idea: Integrate all those integrals by parts a few times before you start differentiating - that way the integrals will be convergent even after differentiating.2012-02-01
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    Which $u$ and $dv$ would you choose?2012-02-01
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    @Peter, for ${\vartheta _2}\left( t \right) = - \frac{\pi }{2}\int\limits_0^\infty {\frac{{\sin tu}}{{{u^2} + 1}}du}$ I would choose to differentiate the $ \frac{1}{u^2+1} $ and integrate the sine term. This will yield stronger powers of growth in the denominator, speeding up the convergence. Then differentiating will still yield a convergent integral. Also, make sure you tag me "@RagibZaman" or else I don't get notified of your comment!2012-02-01
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    @RagibZaman I'd like to see if someone can provide a proof using LT.2012-02-02
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    I guess I've found a solution.2012-02-02
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    @Peter I guess your edit containing a solution would be more appropriate as an answer.2013-05-24
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    @Lord_Farin Do you have any other solution?2013-05-24

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