The Laplace transform of e i5t where i = √−1, is

CONCEPT:From the BASIC definition of Laplace transform:\({\rm{F}}\left( {\rm{s}} \right) = \;\mathop \smallint \limits_0^\infty {\rm{f}}\left( {\rm{t}} \right){{\rm{e}}^{{\rm{ - st}}}}{\rm{dt}}\)f(t) = eat\({\rm{F}}\left( {\rm{s}} \right)= \;\mathop \smallint \limits_0^\infty {{\rm{e}}^{{\rm{ - }}\left( {{\rm{s - a}}} \right){\rm{t}}}}{\rm{dt}}\;\)\({\rm{F}}\left( {\rm{s}} \right){\rm{ = }}\frac{{\rm{1}}}{{{\rm{s - a}}}}\)CALCULATION:Given:a= 5i\({\rm{F}}\left( {\rm{s}} \right){\rm{ = }}\frac{{\rm{1}}}{{{\rm{s - a}}}}\)\({\rm{F}}\left( {\rm{s}} \right){\rm{ = }}\frac{{\rm{1}}}{{{\rm{s - 5i}}}}\)\({\rm{F}}\left( {\rm{s}} \right){\rm{ = }}\frac{{\rm{1}}}{{{\rm{s - 5i}}}}\TIMES \frac {s + 5i}{s + 5i}\)\({\rm{F}}\left( {\rm{s}} \right){\rm{ = }}\frac{{\rm{s +5i}}}{{{\rm{s^2 +25}}}}\)