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Question Number 138742 by mnjuly1970 last updated on 17/Apr/21

$$........nice....calculus...$$$$evaluate::$$$$\mathrm{\Omega }={\int }_0^1\frac{x^{e^{\pi }-1}-x^{e^{\gamma }-1}}{ln\left(\sqrt[3]{x}\right)}dx\stackrel{?}{=}3(\pi -\gamma )$$

Answered by Dwaipayan Shikari last updated on 17/Apr/21

$$3{\int }_0^1\frac{x^{e^{\pi }-1}-x^{e^{\gamma }-1}}{log\left(x\right)}dx=3\chi (e^{\pi }-1,e^{\gamma }-1)=3(\pi -\gamma )$$$$\chi (a,b)={\int }_0^1\frac{x^a-x^b}{log\left(x\right)}dx$$$$\frac{\partial }{\partial a}\chi (a,b)=\frac{1}{a+1}\Rightarrow \chi (a,b)=log(a+1)+C$$$$\chi (b,b)=0\Rightarrow C=-log(b+1)\Rightarrow \chi (a,b)=log\left(\frac{a+1}{b+1}\right)$$

Commented by Ar Brandon last updated on 17/Apr/21

$$\mathrm{Wow}\mathrm{cool}!\mathrm{I}\mathrm{was}\mathrm{supposed}\mathrm{to}\mathrm{remain}\mathrm{at}$$$$\mathrm{the}\mathrm{first}\mathrm{power}.$$

Commented by mnjuly1970 last updated on 17/Apr/21

$$tashakor...mercey..$$

Answered by Ar Brandon last updated on 17/Apr/21

$$\mathrm{\Omega }={\int }_0^1\frac{{\mathrm{x}}^{{\mathrm{e}}^{\pi }-1}-{\mathrm{x}}^{{\mathrm{e}}^{\gamma }-1}}{\ln \left(\sqrt[3]{\mathrm{x}}\right)}\mathrm{dx},\mathrm{u}={\mathrm{x}}^{\frac{1}{3}}\Rightarrow {3\mathrm{u}}^2\mathrm{du}=\mathrm{dx}$$$$={\int }_0^1\frac{{\mathrm{u}}^{3({\mathrm{e}}^{\pi }-1)}-{\mathrm{u}}^{3({\mathrm{e}}^{\gamma }-1)}}{\ln \left(\mathrm{u}\right)}\left({3\mathrm{u}}^2\mathrm{du}\right)=3{\int }_0^1\frac{{\mathrm{u}}^{{3\mathrm{e}}^{\pi }-1}-{\mathrm{u}}^{{3\mathrm{e}}^{\gamma }-1}}{\ln \left(\mathrm{u}\right)}\mathrm{du}$$$$\mathrm{f}\left(\alpha \right)=3{\int }_0^1\frac{{\mathrm{u}}^{{3\mathrm{e}}^{\pi }+\alpha }-{\mathrm{u}}^{{3\mathrm{e}}^{\gamma }+\alpha }}{\ln \left(\mathrm{u}\right)}\mathrm{du}$$$$\mathrm{f}{}^{\prime }\left(\alpha \right)=3{\int }_0^1({\mathrm{u}}^{{3\mathrm{e}}^{\pi }+\alpha }-{\mathrm{u}}^{{3\mathrm{e}}^{\gamma }+\alpha })\mathrm{du}$$$$=3{[\frac{{\mathrm{u}}^{{3\mathrm{e}}^{\pi }+\alpha +1}}{{3\mathrm{e}}^{\pi }+\alpha +1}-\frac{{\mathrm{u}}^{{3\mathrm{e}}^{\gamma }+\alpha +1}}{{3\mathrm{e}}^{\gamma }+\alpha +1}]}_0^1$$$$=3[\frac{1}{{3\mathrm{e}}^{\pi }+\alpha +1}-\frac{1}{{3\mathrm{e}}^{\gamma }+\alpha +1}]$$$$\mathrm{f}\left(\alpha \right)=3\ln ({3\mathrm{e}}^{\pi }+\alpha +1)-3\ln ({3\mathrm{e}}^{\gamma }+\alpha +1)+\mathrm{C}$$

Commented by mnjuly1970 last updated on 17/Apr/21

$$thanksalot...$$

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