- #1
Math100
- 802
- 222
- Homework Statement
- Define the logarithmic integral ## li(x) ## by ## li(x)=\int_{2}^{x}\frac{dt}{\log t} ##, for ## x>2 ##. Prove that ## li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A ## and ## li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}+B ##, for some constants ## A ## and ## B ## that you should determine.
- Relevant Equations
- None.
Proof:
Observe that
\begin{align*}
&li(x)=\int_{2}^{x}\frac{dt}{\log t}\\
&=[\frac{t}{\log t}]_{2}^{x}-\int_{2}^{x}t(\frac{1}{\log t})'dt\\
&=\frac{x}{\log x}-\frac{2}{\log 2}-\int_{2}^{x}t(-\frac{1}{t\log^2 t})dt\\
&=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2}\\
&=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A\\
\end{align*}
where ## A=-\frac{2}{\log 2} ##.
This implies ## li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2} ##.
Since
\begin{align*}
&\int_{2}^{x}\frac{dt}{\log^2 t}=[\frac{t}{\log^2 t}]_{2}^{x}+2\int_{2}^{x}\frac{dt}{\log^3 t}\\
&=\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-\frac{2}{\log^2 2},\\
\end{align*}
it follows that ## li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-\frac{2}{\log 2}-\frac{2}{\log^2 2} ##.
Thus ## li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-(\frac{2}{\log 2}+\frac{2}{\log^2 2}) ##, where ## B=-\frac{2}{\log 2}-\frac{2}{\log^2 2} ##.
Therefore, ## li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A ## and
## li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}+B ##
where ## A=-\frac{2}{\log 2} ## and ## B=-\frac{2}{\log 2}-\frac{2}{\log^2 2} ##.
Observe that
\begin{align*}
&li(x)=\int_{2}^{x}\frac{dt}{\log t}\\
&=[\frac{t}{\log t}]_{2}^{x}-\int_{2}^{x}t(\frac{1}{\log t})'dt\\
&=\frac{x}{\log x}-\frac{2}{\log 2}-\int_{2}^{x}t(-\frac{1}{t\log^2 t})dt\\
&=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2}\\
&=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A\\
\end{align*}
where ## A=-\frac{2}{\log 2} ##.
This implies ## li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}-\frac{2}{\log 2} ##.
Since
\begin{align*}
&\int_{2}^{x}\frac{dt}{\log^2 t}=[\frac{t}{\log^2 t}]_{2}^{x}+2\int_{2}^{x}\frac{dt}{\log^3 t}\\
&=\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-\frac{2}{\log^2 2},\\
\end{align*}
it follows that ## li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-\frac{2}{\log 2}-\frac{2}{\log^2 2} ##.
Thus ## li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}-(\frac{2}{\log 2}+\frac{2}{\log^2 2}) ##, where ## B=-\frac{2}{\log 2}-\frac{2}{\log^2 2} ##.
Therefore, ## li(x)=\frac{x}{\log x}+\int_{2}^{x}\frac{dt}{\log^2 t}+A ## and
## li(x)=\frac{x}{\log x}+\frac{x}{\log^2 x}+2\int_{2}^{x}\frac{dt}{\log^3 t}+B ##
where ## A=-\frac{2}{\log 2} ## and ## B=-\frac{2}{\log 2}-\frac{2}{\log^2 2} ##.
Last edited: