I am trying to verify $$\int_0^\infty te^{-t}\log{t}\,\mathrm dt=1-\gamma,$$ where $\gamma$ is Euler–Mascheroni constant. This problem is exercises 10.2.11(b) in Mathematical methods for physicists, 3rd edition by Arfken. I verified problem (a), $\int_0^\infty e^{-t}\log{t}dt=-\gamma$, by using integration by parts. My question is: (1) Can we solve this integral if we do not know the answer in exercise (a)? (b) I let $u=te^{-t}$ and $dv=\log{t}dt$, but the integral becomes more complex. Any help would be appreciated. Thanks in advance.
Integral representations of the Euler–Mascheroni constant
calculusdefinite integralseuler-mascheroni-constant
Best Answer
Independent of the Gamma Function, what immediately came to my mind was that we can use the fact that $\frac{d}{dx}a^x=a^x\log a$. Let us consider $$I(a) = \int_0^{\infty}t^ae^{-t}dt$$ Differentiating under the Integral Sign gives us $I'(a) = \int_0^{\infty}t^ae^{-t}\log tdt$ which is the required integral with $a =1$. Applying Integration by Parts to $I(a)$, $$I(a)=-e^{-t}t^a|_0^{\infty}+a\int_0^{\infty}t^{a-1}e^{-t}dt=aI(a-1)$$ $$\therefore I(a) =a!$$ Considering the recurrence relation and differentiating both sides with respect to a, $$I'(a)=I(a-1)+aI'(a-1)$$ Dividing by $I(a)$ or $aI(a-1)$, $$\frac{I'(a)}{I(a)}=\frac{I'(a-1)}{I(a-1)}+\frac{1}{a}$$ which gives the connection to the Harmonic Numbers and finally to the Euler-Mascheroni Constant. Hope you can complete the rest.