- #1
NastyAccident
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Homework Statement
[tex]\int^{1}_{0} \frac{ln(x)}{sqrt(x)}[/tex]
Homework Equations
Integration by Parts
Improper Fractions: Type II Integral (discontinuous at a)
The Attempt at a Solution
Now, this might seem odd, but I'm having a devil of a time understanding how this is converging when the limit goes to 0+.
Also, I'm trying to get the hang of LaTeX so, please excuse any improper usage.
[tex]
\int^{1}_{0} \frac{ln(x)}{sqrt(x)}
[/tex]
u = ln(x) dv = 1/sqrt(x)
du = 1/x v = 2sqrt(x)
[tex]2sqrt(x)ln(x)\right|^{1}_{0} - \int^{1}_{0} \frac {(2sqrt(x))}{x}[/tex]
[tex]2sqrt(x)ln(x)\right|^{1}_{0} - 2*\int^{1}_{0} \frac {1}{sqrt(x)}[/tex]
[tex]2sqrt(x)ln(x)\right|^{1}_{0} - 4sqrt(x)\right|^{1}_{0}[/tex]
lim [tex][2sqrt(x)*ln(x) - 4sqrt(x)]^{1}_{t}[/tex]
t->0+
Now, here is where my question lies with why this is converging...
lim 2sqrt(1)*ln(1) - 4sqrt(1) - (2sqrt(t)*ln(t) - 4sqrt(t))
t->0+
When you plug t in, you receive ln(t) which is -infinity... Thus, the integral would be diverging and not converging.
Does plugging the t into a square root prior to a natural log cause the term to then be zero instead of -infinity and thus converging?
Thanks!
NastyAccident