Question Number 56169 by MJS last updated on 11/Mar/19 | ||
$$\underset{−\infty} {\int}^{\mathrm{1}} \left({a}+{b}\mathrm{i}\right)^{{x}} {dx}=? \\ $$ | ||
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19 | ||
$$\int{p}^{{x}} {dx}=\frac{{p}^{{x}} }{{lnp}}+{c} \\ $$$${but}\:{log}_{{e}} \left({a}+{ib}\right)=\frac{\mathrm{1}}{\mathrm{2}}{log}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{i}\left(\mathrm{2}{n}\pi+{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\right) \\ $$$${so} \\ $$$$\mid\left({a}+{ib}\right)^{{x}} \mid_{−\infty} ^{\mathrm{1}} /\left[\frac{\mathrm{1}}{\mathrm{2}}{log}_{{e}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{i}\left(\mathrm{2}{n}\pi+{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\right)\right] \\ $$$$\left({a}+{ib}\right)/\left[\frac{\mathrm{1}}{\mathrm{2}{l}}{log}_{{e}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{i}\left(\mathrm{2}{n}\pi+{tan}^{−\mathrm{1}} \frac{{b}}{{a}}\right)\right. \\ $$$${i}\:{have}\:{rectified}... \\ $$ | ||
Commented by Smail last updated on 11/Mar/19 | ||
$$\int{p}^{{x}} {dx}=\int{e}^{{xlnp}} {dx}=\frac{\mathrm{1}}{{lnp}}{e}^{{xlnp}} +{c} \\ $$$$=\frac{{p}^{{x}} }{{lnp}}+{c}\:\: \\ $$ | ||
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19 | ||
$${yes}\:{sir}\:{you}\:{are}\:{right}... \\ $$ | ||
Commented by MJS last updated on 11/Mar/19 | ||
$$\mathrm{thank}\:\mathrm{you} \\ $$ | ||