Improper integration change of variables

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SUMMARY

The discussion focuses on evaluating improper integrals involving logarithmic functions. The first integral, \(\int _1^{\infty }\frac{1}{x^2}\text{Log}[x]dx\), is shown to equal \(-\int_0^1 \text{Log}[x] \, dx\) using the substitution \(x = \frac{1}{u}\). The second integral, \(\int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx\), is evaluated to be zero by splitting it into two parts and applying the same substitution technique. The conclusion emphasizes the importance of variable substitution in solving these integrals.

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Homework Statement



show

[tex]\int _1^{\infty }\frac{1}{x^2}\text{Log}[x]dx=-\int_0^1 \text{Log}[x] \, dx[/tex]

similarly show

[tex]\int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx = 0[/tex]


The Attempt at a Solution



For the first part a substitution 1/x works.

The second part I cannot do, I thought about

[tex]\int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx+\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx[/tex]

and then trying to maybe show

[tex]\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=-\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx[/tex]

but for now I am not sure what to do.
 
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Try substituting u=1/x in one of your final two integrals.
 
[tex]\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=\int _1^0\frac{1}{u^2+1}\text{Log}<u>du </u>[/tex]

[tex]\int _1^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx+\int _0^1\frac{1}{x^2+1}\text{Log}[x]dx=\int _0^{\infty }\frac{1}{x^2+1}\text{Log}[x]dx=0[/tex]

thanks!
 

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