MHB S6-7.1.79 log integral u substitution

[karush]

$\large{7.R.79} $
$\tiny\text{UHW 242 log integral }$
https://www.physicsforums.com/attachments/5717
$$\begin{align}
\displaystyle
x& = \frac{1}{u} & {u}^{2 }du&={dx }
\end{align} $$
$$I=\int_{0}^{\infty}
\frac
{\ln\left({\frac{1}{u}}\right)}
{1+\frac{1}{{u}^{2 }}}
{u}^{2}
\,du \\
Stuck
$$$\tiny\text{ Surf the Nations math study group}$

 

[alyafey22]

$\large{7.R.79} $
$\tiny\text{UHW 242 log integral }$

$$\begin{align}
\displaystyle
x& = \frac{1}{u} & {u}^{2 }du&={dx }
\end{align} $$
$$I=\int_{0}^{\infty}
\frac
{\ln\left({\frac{1}{u}}\right)}
{1+\frac{1}{{u}^{2 }}}
{u}^{2}
\,du \\
Stuck
$$$\tiny\text{ Surf the Nations math study group}$

You can simplify the integral more and represent it in terms of the original integral.

 

[Prove It]

$\large{7.R.79} $
$\tiny\text{UHW 242 log integral }$

$$\begin{align}
\displaystyle
x& = \frac{1}{u} & {u}^{2 }du&={dx }
\end{align} $$
$$I=\int_{0}^{\infty}
\frac
{\ln\left({\frac{1}{u}}\right)}
{1+\frac{1}{{u}^{2 }}}
{u}^{2}
\,du \\
Stuck
$$$\tiny\text{ Surf the Nations math study group}$

If $\displaystyle \begin{align*} x = \frac{1}{u} \end{align*}$ then $\displaystyle \begin{align*} \mathrm{d}x = -\frac{1}{u^2}\,\mathrm{d}u \end{align*}$, not $\displaystyle \begin{align*} u^2\,\mathrm{d}u \end{align*}$. Now notice that if $\displaystyle \begin{align*} x \to 0^+ \end{align*}$ then $\displaystyle \begin{align*} u \to \infty \end{align*}$ and if $\displaystyle \begin{align*} x \to \infty \end{align*}$ then $\displaystyle \begin{align*} u \to 0 \end{align*}$. This gives

$\displaystyle \begin{align*} \int_0^{\infty}{ \frac{\ln{(x)}}{1 + x^2}\,\mathrm{d}x } &= \int_{\infty}^0{ \frac{\ln{\left( \frac{1}{u} \right) }}{1 + \left( \frac{1}{u} \right) ^2}\,\left( -\frac{1}{u^2} \right) \,\mathrm{d}u } \\ &= \int_0^{\infty}{ \frac{ \ln{ \left( \frac{1}{u} \right) } }{ u^2\,\left( 1 + \frac{1}{u^2} \right) } \,\mathrm{d}u } \\ &= \int_0^{\infty}{ \frac{\ln{ \left( \frac{1}{u} \right) }}{1 + u^2}\,\mathrm{d}u } \end{align*}$

Can you see a relationship between the two integrals?

 

[karush]

Are you referring to
$$\int_b^a f(x) \ dx = -\int_a^b f(x) \ dx $$

Forget what this called nor do I know why

 

[Prove It]

Are you referring to
$$\int_b^a f(x) \ dx = -\int_a^b f(x) \ dx $$

Forget what this called nor do I know why

Yes I did use that rule.

Anyway, to finish off the problem, would you agree that $\displaystyle \begin{align*} \int_0^{\infty}{ \frac{\ln{\left( \frac{1}{u} \right)}}{1+u^2}\,\mathrm{d}u} \end{align*}$ will have exactly the same value as $\displaystyle \begin{align*} \int_0^{\infty}{ \frac{\ln{ \left( \frac{1}{x} \right) }}{1 + x^2}\,\mathrm{d}x } \end{align*}$? All that's happened is there is a different letter being used in the place of the variable.

So from our original equation $\displaystyle \begin{align*} \int_0^{\infty}{ \frac{\ln{(x)}}{1 + x^2}\,\mathrm{d}x } = \int_0^{\infty}{ \frac{\ln{ \left( \frac{1}{u} \right) }}{1 + u^2}\,\mathrm{d}u } \end{align*}$ we can write $\displaystyle \begin{align*} \int_0^{\infty}{ \frac{\ln{(x)}}{1 + x^2}\,\mathrm{d}x } = \int_0^{\infty}{ \frac{\ln{ \left( \frac{1}{x} \right) }}{1 + x^2}\,\mathrm{d}x } \end{align*}$.

Now since $\displaystyle \begin{align*} \ln{ \left( \frac{1}{x} \right) } = \ln{ \left( x^{-1} \right) } = -\ln{(x)} \end{align*}$ that means we have $\displaystyle \begin{align*} \int_0^{\infty}{ \frac{\ln{(x)}}{1 + x^2}\,\mathrm{d}x } = -\int_0^{\infty}{ \frac{\ln{(x)}}{1 + x^2}\,\mathrm{d}x } \end{align*}$ and so the final result should now be obvious.

 

[alyafey22]

There is another approach to see that the integral goes to 0. Set $x=e^{-t}$. And realize the function generated is odd.

 

[karush]

There is another approach to see that the integral goes to 0. Set $x=e^{-t}$. And realize the function generated is odd.

I really appreciate all the help with this
the class starts August 22.
so feeling more confident I will do well
due to this forum.