Integral representation for Euler constant

Question

I am trying to show that $$\gamma=\int_0^{\infty}\frac{\ln(1+x)+e^{-x}-1}{x^2}\;dx$$ I haven't done so much, but since this is related to $\psi (1)=\Gamma'(1)$ I have tried to work backwards. So $$\Gamma(x)=\int_0^{\infty} t^{x-1}e^{-t}\;dt\rightarrow \Gamma'(x)=\int_0^{\infty} t^{x-1}e^{-t}\ln t\;dt$$ $$\int_0^{1} t^{x-1}e^{-t}\ln t\;dt+\int_0^{1} \frac{e^{-\frac{1}{t}}\ln t}{t^2}\;dt$$ Where I used $t=\frac{1}{t}$ in the second integral, I also tried to recombine those two, but of no use. Also this is equal to: $$\int_0^{\infty} (\ln\left(1+\frac{1}{x}\right)+e^{-\frac{1}{x}}-1)dx$$ Now I am thinking to use series. Could you give me some help or hints?

Answer 1

Taking integration by parts (IbP) twice, we obtain

\begin{align*} \int_{0}^{\infty} \frac{\log(1+x) + e^{-x} - 1}{x^2} \, dx &\stackrel{\text{(IbP)}}{=} \int_{0}^{\infty} \left( \frac{1}{x+1} - e^{-x} \right) \frac{1}{x} \, dx \\ &\stackrel{\text{(IbP)}}{=} \int_{0}^{\infty} \left( \frac{1}{(x+1)^2} - e^{-x} \right) \log x \, dx. \end{align*}

But it is easy to check that

$$ \int_{0}^{\infty} \frac{\log x}{(1+x)^2} \, dx \stackrel{(x\ \mapsto\ \frac{1}{x})}{=} -\int_{0}^{\infty} \frac{\log x}{(1+x)^2} \, dx $$

and hence the common value of both sides is zero. For the remaining term, it is well-known that

$$ \int_{0}^{\infty} e^{-x}\log x \, dx = -\gamma. $$

This particular integral has been explained several times in this community. (See this, for instance.)

Answer 2

Hint:

we have that $$ \eqalign{ & I(a,b) = \int_{\,x\, = \,a}^{\;b} {{{\ln \left( {1 + x} \right) + e^{\, - x} - 1} \over {x^{\,2} }}dx} = \cr & = - \int_{\,x\, = \,a}^{\;b} {\ln \left( {1 + x} \right)d\left( {1/x} \right)} + \int_{\,x\, = \,a}^{\;b} {{{e^{\, - x} - 1} \over {x^{\,2} }}dx} = \cr & = \int_{\,y\, = \,1/b}^{\;1/a} {\ln \left( {1 + 1/y} \right)dy} + \int_{\,x\, = \,a}^{\;b} {{{e^{\, - x} - 1} \over {x^{\,2} }}dx} = \cr & = \int_{\,y\, = \,1/b}^{\;1/a} {\left( {\ln \left( {1 + y} \right) - \ln y} \right)dy} + \left. {\left( {E_1 (x) - {{e^{\, - x} } \over x}} \right)\,} \right|_{\,x\, = \,a}^{\;b} = \cr & \left. { = \left( {\left( {1 + y} \right)\ln \left( {1 + y} \right) - y\ln y} \right)\,} \right|_{\,y\, = \,1/b}^{\;1/a} + \left. {\left( {E_1 (x) - {{e^{\, - x} } \over x}} \right)\,} \right|_{\,x\, = \,a}^{\;b} = \cr & \quad \cdots \cr} $$

and in fact the Exponential integral function is tied to $\gamma$.