Question Number 27183 by abdo imad last updated on 02/Jan/18
$${find}\:{the}\:{value}\:{of}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}−\mathrm{1}}{{lnt}}{dt}\:. \\ $$
Answered by prakash jain last updated on 03/Jan/18
![F(x)=∫_0 ^1 ((t^x (t−1))/(ln t))dt (dF/dx)=∫_0 ^1 (∂/∂x) (((t^x (t−1))/(ln t)))dt (dF/dx)=∫_0 ^1 ((t^x (t−1))/(ln t))ln t dt (dF/dx)=∫_0 ^1 t^(x+1) −t^x dt=[(t^(x+2) /(x+2))−(t^(x+1) /(x+1))]_0 ^1 =(1/(x+2))−(1/(x+1)) F(x)=ln ∣((x+2)/(x+1))∣+C F(−∞)=0⇒C=0 F(x)=ln ∣((x+2)/(x+1))∣ I=F(0)=ln 2](https://www.tinkutara.com/question/Q27238.png)
$${F}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{x}} \left({t}−\mathrm{1}\right)}{\mathrm{ln}\:{t}}{dt} \\ $$$$\frac{{dF}}{{dx}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\partial}{\partial{x}}\:\left(\frac{{t}^{{x}} \left({t}−\mathrm{1}\right)}{\mathrm{ln}\:{t}}\right){dt} \\ $$$$\frac{{dF}}{{dx}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{x}} \left({t}−\mathrm{1}\right)}{\mathrm{ln}\:{t}}\mathrm{ln}\:{t}\:{dt} \\ $$$$\frac{{dF}}{{dx}}=\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{{x}+\mathrm{1}} −{t}^{{x}} \:{dt}=\left[\frac{{t}^{{x}+\mathrm{2}} }{{x}+\mathrm{2}}−\frac{{t}^{{x}+\mathrm{1}} }{{x}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{x}+\mathrm{2}}−\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$${F}\left({x}\right)=\mathrm{ln}\:\mid\frac{{x}+\mathrm{2}}{{x}+\mathrm{1}}\mid+{C} \\ $$$${F}\left(−\infty\right)=\mathrm{0}\Rightarrow{C}=\mathrm{0} \\ $$$${F}\left({x}\right)=\mathrm{ln}\:\mid\frac{{x}+\mathrm{2}}{{x}+\mathrm{1}}\mid \\ $$$${I}={F}\left(\mathrm{0}\right)=\mathrm{ln}\:\mathrm{2} \\ $$
Commented by prakash jain last updated on 03/Jan/18
$$\mathrm{please}\:\mathrm{comment}\:\mathrm{if}\:\mathrm{mistakes}.\:\mathrm{thanks} \\ $$