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Question Number 63510 by turbo msup by abdo last updated on 05/Jul/19

$$letf\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{x+t}dtwithx>0$$$$and0<a<1$$$$1)calculatef\left(x\right)$$$$2)calculateg\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(x+t)}^2}dt$$$$3)findthevalueof{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(1+t)}^2}dt$$

Commented by mathmax by abdo last updated on 06/Jul/19

$$1)changementt=xugivef\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{{\left(xu\right)}^{a-1}}{x+xu}xdu=x^{a-1}{\int }_0^{\mathrm{\infty }}\frac{u^{a-1}}{1+u}du$$$$=x^{a-1}\frac{\pi }{sin\left(\pi a\right)}\Rightarrow f\left(x\right)=\frac{\pi x^{a-1}}{sin\left(\pi a\right)}$$$$2)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{\partial }{\partial x}\left(\frac{t^{a-1}}{x+t}\right)dt=-{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(x+t)}^2}dt=-g\left(x\right)\Rightarrow$$$$g\left(x\right)=-f^{{}^{\prime }}\left(x\right)butf\left(x\right)=\frac{\pi }{sin\left(\pi a\right)}{e}^{(a-1)ln\left(x\right)}\Rightarrow$$$$f^{{}^{\prime }}\left(x\right)=\frac{\pi (a-1)}{xsin\left(\pi a\right)}{x}^{a-1}=\frac{\pi (a-1)x^{a-2}}{sin\left(\pi a\right)}\Rightarrow g\left(x\right)=\frac{\pi (1-a)}{sin\left(\pi a\right)}{x}^{a-2}$$$$3){\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(1+t)}^2}dt=g\left(1\right)=\frac{\pi (1-a)}{sin\left(\pi a\right)}(0<a<1).$$$$remarkwehaveforallintegrn$$$$f^{\left(n\right)}\left(x\right)={\int }_0^{\mathrm{\infty }}\frac{{(-1)}^{n}n!t^{a-1}}{{(x+t)}^{n+1}}dt\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(x+t)}^{n+1}}dt=\frac{{(-1)}^n}{n!}{f}^{\left(n\right)}\left(x\right)$$$$wehavef\left(x\right)=\frac{\pi }{sin\left(\pi a\right)}{e}^{(a-1)ln\left(x\right)}letfind{\left(e^{\lambda ln\left(x\right)}\right)}^{\left(n\right)}$$$${\left(e^{\lambda lnx}\right)}^{\left(1\right)}=\frac{\lambda }{x}{e}^{\lambda ln\left(x\right)}\Rightarrow {\left(e^{\lambda lnx}\right)}^{\left(2\right)}=\frac{{\lambda }^2}{x^2}{e}^{\lambda ln\left(x\right)}\Rightarrow {\left(e^{\lambda ln\left(x\right)}\right)}^{\left(n\right)}=\frac{{\lambda }^n}{x^n}{e}^{\lambda lnx}\Rightarrow$$$$f^{\left(n\right)}\left(x\right)=\frac{\pi }{sin\left(\pi a\right)}\frac{{(a-1)}^n}{x^n}{x}^{a-1}=\frac{\pi {(a-1)}^n}{sin\left(\pi a\right)}{x}^{a-n-1}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(x+t)}^{n+1}}dt=\frac{\pi {(1-a)}^n}{n!sin\left(\pi a\right)}{x}^{a-n-1}$$$$specialcasex=1\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(1+t)}^{n+1}}dt=\frac{\pi {(1-a)}^n}{n!sin\left(\pi a\right)}.$$$$$$

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