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Question Number 83385 by M±th+et£s last updated on 01/Mar/20

$$\int \frac{1+{ae}^{-(a+1)x}+(a+1)e^{-ax}}{x^2}dx,a\in z^{+}$$

Commented by M±th+et£s last updated on 01/Mar/20

$$typo$$$${\int }_0^{\mathrm{\infty }}$$

Answered by mind is power last updated on 01/Mar/20

$${\int }_0^{+\mathrm{\infty }}\frac{1+{ae}^{-(a+1)x}-(a+1)e^{-ax}}{x^2}dx?$$

Commented by M±th+et£s last updated on 01/Mar/20

$$yessir$$

Commented by mathmax by abdo last updated on 02/Mar/20

$$byparts{u}^{{}^{\prime }}=\frac{1}{x^2}andv=1+{ae}^{-(a+1)x}-(a+1)e^{-ax}\Rightarrow$$$$I={[-\frac{1}{x}(1+{ae}^{-(a+1)x}-(a+1)e^{-ax})]}_0^{+\mathrm{\infty }}+$$$${\int }_0^{\mathrm{\infty }}(e^{-(a+1)x}-a(a+1)e^{-(a+1)x}-e^{-ax}+a(a+1)e^{-ax})\frac{dx}{x}$$$$={\int }_0^{\mathrm{\infty }}\frac{e^{-(a+1)x}-e^{-ax}}{x}dx-a(a+1){\int }_0^{\mathrm{\infty }}\frac{e^{-(a+1)x}-e^{-ax}}{x}dx$$$$=(1-a^2-a){\int }_0^{\mathrm{\infty }}\frac{e^{-(a+1)x}-e^{-ax}}{x}dxlet\xi \left(a\right)={\int }_0^{\mathrm{\infty }}\frac{e^{-(a+1)x}-e^{-ax}}{x}dx$$$$\Rightarrow {\xi }^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{-x{e}^{-(a+1)x}+{xe}^{-ax}}{x}dx$$$$={\int }_0^{\mathrm{\infty }}{e}^{-ax}dx-{\int }_0^{\mathrm{\infty }}{e}^{-(a+1)x}dx={[-\frac{e^{-ax}}{a}]}_0^{\mathrm{\infty }}-{[-\frac{e^{-(a+1)x}}{a+1}]}_0^{+\mathrm{\infty }}$$$$=\frac{1}{a}-\frac{1}{a+1}\Rightarrow \xi \left(a\right)=ln\left(\frac{a}{a+1}\right)+c(wesupoosea>0)$$$${lim}_{a\to +\mathrm{\infty }}\xi \left(a\right)=0=c\Rightarrow \xi \left(a\right)=ln\left(\frac{a}{a+1}\right)\Rightarrow I=(1-a^2-a)ln\left(\frac{a}{a+1}\right)$$$$resttoprovethat{lim}_{x\to 0}\frac{1+{ae}^{-(a+1)x}-(a+1)e^{-ax}}{x}=0$$$$e^{-(a+1)x}\sim 1-(a+1)x+\frac{(a+1)x^2}{2}\Rightarrow a{e}^{-(a+1)x}\sim a-a(a+1)x+\frac{a(a+1)x^2}{2}$$$$e^{-ax}\sim 1-ax+\frac{a^2{x}^2}{2}\Rightarrow (a+1)e^{-ax}\sim a+1-a(a+1)x+\frac{a^2(a+1)x^2}{2}\Rightarrow$$$$(...-(....)\sim 1+a-a(a+1)x+\frac{a(a+1)x^2}{2}-a-1-a(a+1)x-\frac{a^2(a+1)x^2}{2}$$$$=\frac{a+1}{2}(a-a^2)x^2\Rightarrow \frac{(...)-(...)}{x}\sim \frac{(a+1)(a-a^2)x}{2}\to 0(x\to 0)$$

Answered by mind is power last updated on 02/Mar/20

$$f\left(t\right)={\int }_0^{+\mathrm{\infty }}\frac{1+{\mathrm{ae}}^{-(\mathrm{a}+1)t\mathrm{x}}-(\mathrm{a}+1){\mathrm{e}}^{-\mathrm{a}t\mathrm{x}}}{{\mathrm{x}}^2}$$$$f^{\prime }\left(t\right)={\int }_0^{+\mathrm{\infty }}\frac{a(a+1)(-e^{-(a+1)tx}+e^{-atx})}{x}dx$$$${\int }_0^{+\mathrm{\infty }}\frac{e^{-atx}-e^{-(a+1)tx}}{x}dx=ln\left(\frac{a+1}{a}\right)$$$$f^{\prime }\left(t\right)=a(a+1)ln\left(\frac{a+1}{a}\right)$$$$f\left(t\right)=a(a+1)ln\left(\frac{a+1}{\mathrm{a}}\right)\mathrm{t}+\mathrm{c}$$$$f\left(0\right)=0\Rightarrow c=0$$$$f\left(t\right)=a(a+1)ln\left(\frac{1+a}{a}\right)t$$$$\int \frac{1+{ae}^{-(a+1)x}-(a+1)e^{-ax}}{x^2}dx=f\left(1\right)$$$$=a(a+1)ln\left(\frac{1+a}{a}\right)$$$$$$

Commented by mind is power last updated on 02/Mar/20

$$yeahsorryverrybusythesedayworck+studies$$$$f\left(t\right)={\int }_0^{+\mathrm{\infty }}\frac{1+{ae}^{-t(a+1)x}-(1+a)e^{-tax}}{x^2}dx$$$$f\left(0\right)=0\Rightarrow c=0yes$$

Commented by M±th+et£s last updated on 02/Mar/20

$$godblessyousirandthankforthesolution$$

Commented by M±th+et£s last updated on 02/Mar/20

$$thankyousirbutithinkthereissome$$$$typos$$

Commented by M±th+et£s last updated on 02/Mar/20

Commented by M±th+et£s last updated on 02/Mar/20

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