Sk Ramis

Abdullah AL Shohag

${\displaystyle \int _{-2}^{2}(x^{3}\cos \left({\frac {x}{2}}\right)+{\tfrac {1}{2}}){\sqrt {4-x^{2}}}dx}$

"An equation for me has no meaning, unless it expresses a thought of God" - S. Ramanujan

Defining negative value of logarithms

${\displaystyle y=\log(-x)}$

${\displaystyle \Longrightarrow 10^{y}=-x}$

${\displaystyle \Longrightarrow 10^{a+ib}=-x}$${\displaystyle \Longrightarrow 10^{a}.10^{ib}=-x}$

${\displaystyle \Longrightarrow 10^{a}.e^{ib.\ln 10}=-x}$

${\displaystyle \Longrightarrow 10^{a}.{\biggl (}\cos b+i\sin b{\biggr )}^{\ln 10}=-x}$

If we consider ${\displaystyle i\sin b=0}$, then ${\displaystyle b=n\pi }$ but ${\displaystyle \cos b}$ will be negative only when ${\displaystyle b=(2n-1)\pi }$ where ${\displaystyle n\in \mathbb {N} }$ (for simplicity we will take ${\displaystyle n=1}$, so ${\displaystyle b=\pi }$)

${\displaystyle \Longrightarrow 10^{a}.(-1)^{\ln 10}=-x}$

${\displaystyle \Longrightarrow a+{\ln 10}.log(-1)=\log(-x)}$     

${\displaystyle \Longrightarrow a+{\ln 10}.{\log e}.{i\pi }={\log e}.{\ln(-x)}}$ /* ${\displaystyle \log(-1)=i\pi \log e}$ */

${\displaystyle \Longrightarrow a+{\ln 10}.{\log e}.{i\pi }={\log e}.{\biggl (}{\ln(x)}+{i\pi }{\biggr )}}$

${\displaystyle \Longrightarrow a={\log e}.{\ln x}+{i\pi }{\biggl (}1-{\ln 10}{\biggr )}{\log e}}$

if we put the value of ${\displaystyle a}$ in the marked line,

${\displaystyle \Longrightarrow {\log e}.{\ln x}+{i\pi }{\biggl (}1-{\ln 10}{\biggr )}{\log e}+{\ln 10}.log(-1)=\log(-x)}$

${\displaystyle \Longrightarrow {\log x}+{i\pi }{\log e}=\log(-x)}$

so if we generalize it for any base, we get,

${\displaystyle {\log _{a}x}+{i\pi }{\log _{a}e}=\log _{a}(-x)}$, where ${\displaystyle a>1}$ commons:User:ShohagS