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1-x-i-lnxdx-i-2-1-

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Question Number 148376 by ArielVyny last updated on 27/Jul/21
∫_1 ^∞ x^i lnxdx i^2 =−1
$$\int_{\mathrm{1}} ^{\infty} {x}^{{i}} {lnxdx}\:\:\:\:\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 27/Jul/21
I=∫_1 ^∞ x^i logx dx cha7gement logx=t give x=e^t ⇒ I=∫_0 ^∞ (e^t )^i t e^(t ) dt=∫_0 ^∞ t e^(it+t) dt =∫_0 ^∞ t e^((1+i)t) dt =_((1+i)t=z) ∫_0 ^∞ (z/(1+i))e^z (dz/(1+i))=(1/((1+i)^2 ))∫_0 ^∞ ze^z dz by parts but ∫_0 ^∞ ze^(z ) dz is divergent...!
$$\mathrm{I}=\int_{\mathrm{1}} ^{\infty} \:\mathrm{x}^{\mathrm{i}} \mathrm{logx}\:\mathrm{dx}\:\mathrm{cha7gement}\:\mathrm{logx}=\mathrm{t}\:\mathrm{give}\:\mathrm{x}=\mathrm{e}^{\mathrm{t}} \:\Rightarrow \\ $$$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{e}^{\mathrm{t}} \right)^{\mathrm{i}} \mathrm{t}\:\mathrm{e}^{\mathrm{t}\:} \mathrm{dt}=\int_{\mathrm{0}} ^{\infty} \mathrm{t}\:\mathrm{e}^{\mathrm{it}+\mathrm{t}} \:\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{t}\:\mathrm{e}^{\left(\mathrm{1}+\mathrm{i}\right)\mathrm{t}} \:\mathrm{dt} \\ $$$$=_{\left(\mathrm{1}+\mathrm{i}\right)\mathrm{t}=\mathrm{z}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}}{\mathrm{1}+\mathrm{i}}\mathrm{e}^{\mathrm{z}} \:\frac{\mathrm{dz}}{\mathrm{1}+\mathrm{i}}=\frac{\mathrm{1}}{\left(\mathrm{1}+\mathrm{i}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\mathrm{ze}^{\mathrm{z}} \:\mathrm{dz}\:\mathrm{by}\:\mathrm{parts} \\ $$$$\mathrm{but}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{ze}^{\mathrm{z}\:} \:\mathrm{dz}\:\mathrm{is}\:\mathrm{divergent}…! \\ $$
Commented by ArielVyny last updated on 28/Jul/21
thank sir
$${thank}\:{sir} \\ $$

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