# 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

$$- \frac{1}{{{s^2} - 1}}\int\limits_0^\infty {\frac{{\log m + \log s}}{{{m^2} + 1}}} dm = - \frac{{\log s}}{{{s^2} - 1}}\int\limits_0^\infty {\frac{1}{{{m^2} + 1}}} dm = - \frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$

-
The integral you have doesn't converge. – anon Feb 1 '12 at 2:54
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. – Ragib Zaman Feb 1 '12 at 2:59
Which $u$ and $dv$ would you choose? – Pedro Tamaroff Feb 1 '12 at 4:07
@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! – Ragib Zaman Feb 1 '12 at 9:30
@RagibZaman I'd like to see if someone can provide a proof using LT. – Pedro Tamaroff Feb 2 '12 at 0:45

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

$$- \frac{\pi }{4}\int\limits_0^\infty {\frac{{dm}}{{{s^2} + m}}\frac{1}{{m + 1}}}$$

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

$$- \frac{1}{{{s^2} - 1}}\int\limits_0^\infty {\frac{{\log m + \log s}}{{{m^2} + 1}}} dm = - \frac{{\log s}}{{{s^2} - 1}}\int\limits_0^\infty {\frac{1}{{{m^2} + 1}}} dm = - \frac{\pi }{2}\frac{{\log s}}{{{s^2} - 1}}$$

-