I need a confirmation and answer about the following problem:
If we have $ g(x)=\ln x+{ x }^{ -1/2 }{ 1 }_{ x\le 1 } $
I'm trying to determine $ \int _{ 0 }^{ +\infty }{ \ln x+{ x }^{ -1/2 }{ 1 }_{ x\le 1 } } $
$$ \int _{ 0 }^{ +\infty }{ \ln x+{ x }^{ -1/2 }{ 1 }_{ x\le 1 } } =\int _{ 0 }^{ 1 }{ \ln x+{ x }^{ -1/2 }{ 1 }_{ x\le 1 } } +\int _{ 1 }^{ +\infty }{ \ln x } $$
For the first part:
$$\begin{align} \int _{ 0 }^{ 1 }{ \ln x+{ x }^{ -1/2 }{ 1 }_{ x\le 1 } } & ={ \left[ x\ln x-x+2\sqrt { x } \right] }_{ 0 }^{ 1 }\\ & =1 \ln 1-1+2\sqrt { 1 } -X \ln X-0+2\sqrt { 0 } \\ & =1-X \ln X \end{align}$$
which tends to $1$ at $0$.
For the second part:
$$\begin{align} \int _{ 1 }^{ +\infty }{ \ln x } ={ \left[ x \ln x-x \right] }_{ 1 }^{ +\infty } & =X \ln X-X-1 \ln 1-1 \\ & =X \ln X-X-1 \end{align}$$
This thing tends to $+\infty$ as $X$ grows to $+\infty$.
So for me the whole $\int _{ 0 }^{ +\infty }{ \ln x+{ x }^{ -1/2 }{ 1 }_{ x\le 1 } }=+\infty$
- Am I correct ?
- Does it mean that this integral is not defined ? I'm not sure what it means to say that an integral is not defined.
Thanks in advance please.