Question Number 27692 by abdo imad last updated on 12/Jan/18
![find by two ways the value of ∫∫_([0,1]) x^y dxdxy then calculate ∫_0 ^1 ((t−1)/(lnt))dt .](https://www.tinkutara.com/question/Q27692.png)
$${find}\:{by}\:{two}\:{ways}\:{the}\:{value}\:{of}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]} \:\:{x}^{{y}} \:\:{dxdxy}\:{then} \\ $$$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}−\mathrm{1}}{{lnt}}{dt}\:\:. \\ $$
Commented by abdo imad last updated on 13/Jan/18
![let put I= ∫∫_([0,1]^2 ) x^y dxdy I= ∫_0 ^1 (∫_0 ^1 ( e^(ylnx) dy)dx but ∫_0 ^1 e^(ylnx) dy = [(1/(lnx)) x^y ]_(y=0) ^(y=1) = ((x−1)/(lnx)) ⇒ I= ∫_0 ^1 ((x−1)/(lnx))dx let change the integral due to fubini theorem I=∫_0 ^1 ( ∫_0 ^1 x^y dx)dy but ∫_0 ^1 x^y dx =[ (x^(y+1) /(y+1))]_(x=0) ^(x=1) = (1/(y+1)) ⇒I= ∫_0 ^1 (dy/(y+1)) =[ln/y+1/]_0 ^1 =ln(2).finally ∫_0 ^1 ((x−1)/(lnx))dx= ln(2).](https://www.tinkutara.com/question/Q27738.png)
$${let}\:{put}\:{I}=\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]^{\mathrm{2}} } \:{x}^{{y}} \:{dxdy} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \left(\:{e}^{{ylnx}} {dy}\right){dx}\:\:{but}\right. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{ylnx}} {dy}\:=\:\:\:\left[\frac{\mathrm{1}}{{lnx}}\:{x}^{{y}} \:\:\right]_{{y}=\mathrm{0}} ^{{y}=\mathrm{1}} \:\:=\:\frac{{x}−\mathrm{1}}{{lnx}}\:\Rightarrow\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}−\mathrm{1}}{{lnx}}{dx} \\ $$$${let}\:{change}\:{the}\:{integral}\:{due}\:{to}\:{fubini}\:{theorem} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{y}} {dx}\right){dy}\:\:{but}\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{y}} {dx}\:=\left[\:\frac{{x}^{{y}+\mathrm{1}} }{{y}+\mathrm{1}}\overset{{x}=\mathrm{1}} {\right]}_{{x}=\mathrm{0}} \\ $$$$=\:\frac{\mathrm{1}}{{y}+\mathrm{1}}\:\:\Rightarrow{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dy}}{{y}+\mathrm{1}}\:=\left[{ln}/{y}+\mathrm{1}/\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:={ln}\left(\mathrm{2}\right).{finally} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}−\mathrm{1}}{{lnx}}{dx}=\:{ln}\left(\mathrm{2}\right). \\ $$