Integral representations of the Euler–Mascheroni constant

Question

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.

Answer 1

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.

Answer 2

That integral is $$ \Gamma'(2)=\Gamma(2)\psi(2)=\psi(2)=H_1-\gamma=1-\gamma $$

Answer 3

Let $ x>0 $, $ t\mapsto\mathrm{e}^{-xt}\ln{t} $ is integrable over $ \mathbb{R}_{+}^{*}=\left(0,+\infty\right) $, and by substituting $ \small\left\lbrace\begin{aligned}y&=xt\\ \mathrm{d}t &=\frac{\mathrm{d}y}{x}\end{aligned}\right. $, we get : \begin{aligned} \int_{0}^{+\infty}{\mathrm{e}^{-xt}\ln{t}\,\mathrm{d}t}=\frac{1}{x}\int_{0}^{+\infty}{\mathrm{e}^{-y}\ln{\left(\frac{y}{x}\right)}\,\mathrm{d}y}&=\frac{1}{x}\int_{0}^{+\infty}{\mathrm{e}^{-y}\ln{y}\,\mathrm{d}y}-\frac{\ln{x}}{x}\int_{0}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y}\\ &=\frac{1}{x}\int_{0}^{+\infty}{\mathrm{e}^{-y}\ln{y}\,\mathrm{d}y}-\frac{\ln{x}}{x} \end{aligned}

Now working on $ \int_{0}^{+\infty}{\mathrm{e}^{-y}\,\mathrm{d}y} $ :

Let $ n $ be a positive integer :

\begin{aligned}\int_{0}^{1}{x^{n-1}\ln{\left(1-x\right)}\,\mathrm{d}x}&=\left[\frac{x^{n}-1}{n}\ln{\left(1-x\right)}\right]_{0}^{1}-\frac{1}{n}\int_{0}^{1}{\frac{1-x^{n}}{1-x}\,\mathrm{d}x}\\ &=-\frac{1}{n}\int_{0}^{1}{\left(\sum_{k=0}^{n-1}{x^{k}}\right)\mathrm{d}x}\\ \int_{0}^{1}{x^{n-1}\ln{\left(1-x\right)}\,\mathrm{d}x}&=-\frac{1}{n}\sum_{k=1}^{n}{\frac{1}{k}}\end{aligned}

Thus, for every positive integer $ n $, substituting $ \small\left\lbrace\begin{aligned}y&=1-\frac{x}{n}\\ \mathrm{d}x &=-n\,\mathrm{d}y\end{aligned}\right. $, will give us : \begin{aligned}\int_{0}^{n}{\left(1-\frac{x}{n}\right)^{n}\ln{x}\,\mathrm{d}x}&=-n\int_{0}^{1}{y^{n}\ln{\left(n\left(1-y\right)\right)}\,\mathrm{d}y}\\ &=-n\ln{n}\int_{0}^{1}{y^{n}\,\mathrm{d}y}-n\int_{0}^{1}{y^{n}\ln{\left(1-y\right)}\,\mathrm{d}y}\\ \int_{0}^{n}{\left(1-\frac{x}{n}\right)^{n}\ln{x}\,\mathrm{d}x}&=-\frac{n}{n+1}\left(\sum_{k=1}^{n}{\frac{1}{k}}-\ln{n}\right)\end{aligned}

Thus : $$ \lim_{n\to +\infty}{\int_{0}^{n}{\left(1-\frac{x}{n}\right)^{n}\ln{x}\,\mathrm{d}x}}=-\gamma $$

By applying the dominated convergence theorem on a function $ f_{n} $ defined on $ \mathbb{R}_{+} $ as follows : $$ \left(\forall x\in\mathbb{R}_{+}\right),\ f_{n}\left(x\right)=\left\lbrace\begin{aligned}\left(1-\frac{x}{n}\right)^{n}\ln{x}\ \ \ \ \ \ \ & \textrm{If }0\leq x\leq n\\ 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ &\textrm{If }\ \ \ \ x\geq n\end{aligned}\right. $$We get that : $$ -\gamma =\small\lim_{n\to +\infty}{\int_{0}^{n}{\left(1-\frac{x}{n}\right)^{n}\ln{x}\,\mathrm{d}x}}=\lim_{n\to +\infty}{\int_{0}^{+\infty}{f_{n}\left(x\right)\mathrm{d}x}}=\int_{0}^{+\infty}{\lim_{n\to +\infty}{f_{n}\left(x\right)}\,\mathrm{d}x}=\int_{0}^{+\infty}{\mathrm{e}^{-x}\ln{x}\,\mathrm{d}x} $$

Hence, $ \forall x>0 $, we have : $$ \int_{0}^{+\infty}{\mathrm{e}^{-xt}\ln{t}\,\mathrm{d}t}=-\frac{\gamma +\ln{x}}{x} $$

Differentiating both sides then taking $x=1 $, we get : $$ \int_{0}^{+\infty}{t\,\mathrm{e}^{-t}\ln{t}\,\mathrm{d}t}=1-\gamma $$