# What is $\lim_{\epsilon\rightarrow0}\int_{\epsilon}^{\infty}\frac{e^{-u}-e^{-u^{\alpha}x}}{u}~\mathrm{d}u$?

I need the integral

$$f(x)=\lim_{\epsilon\rightarrow0}\int_{\epsilon}^{\infty}\frac{\exp(-u)-\exp(-u^{\alpha}x)}{u}~\mathrm{d}u$$

with $$\alpha>0$$ in (statistical learning) application. It looks a lot like

$$\ln(x)=\lim_{\epsilon\rightarrow0}\int_{\epsilon}^{\infty}\frac{\exp(-u)-\exp(-ux)}{u}~\mathrm{d}u$$

which is just $$\alpha=1$$.

Using

$$I_{u}(x)= \int_{1}^{x}e^{-u\chi}~\mathrm{d}\chi=\left[-\frac{1}{u}e^{-u\chi}\right]_{1}^{x}=-\frac{e^{-ux}-e^{-u}}{u}=\frac{e^{-u}-e^{-ux}}{u},$$

I can prove the case for $$\alpha=1$$ by

$$\int_{\epsilon}^{\infty}I_{u}(x)~\mathrm{d}u =\lim_{\epsilon\rightarrow0}\int_{\epsilon}^{\infty}\int_{1}^{x}e^{-u\chi}~\mathrm{d}\chi~\mathrm{d}u=\lim_{\epsilon\rightarrow0}\int_{1}^{x}\int_{\epsilon}^{\infty}e^{-u\chi}~\mathrm{d}u~\mathrm{d}\chi \\ =\lim_{\epsilon\rightarrow0}\int_{1}^{x}\left[-\frac{1}{\chi}e^{-u\chi}\right]_{\epsilon}^{\infty}~\mathrm{d}\chi=\lim_{\epsilon\rightarrow0}\int_{1}^{x}\left(\frac{1}{\chi}e^{-\epsilon\chi}\right)~\mathrm{d}\chi\\ =\lim_{\epsilon\rightarrow0}\int_{1}^{x}\left(\frac{1}{\chi}e^{-\epsilon\chi}\right)~\mathrm{d}\chi=\int_{1}^{x}\left(\lim_{\epsilon\rightarrow0}\frac{1}{\chi}e^{-\epsilon\chi}\right)~\mathrm{d}\chi=\int_{1}^{x}\left(\frac{1}{\chi}\right)~\mathrm{d}\chi=\ln x$$

but how does that transfer?

$$f'(x)=\frac{\text{d} f(x)}{\text{d}x}=\int_{0}^{\infty} \frac{\partial }{\partial x}\left(\frac{e^{-u}-e^{-u^{\alpha}x}}{u}\right) du$$

Taking, $$z=u^{\alpha}x$$ we get

$$=\frac{x^{-1}}{\alpha}(\int_{0}^{\infty} e^{-z} dz) =\frac{x^{-1}}{\alpha}$$.....(1)

Taking, $$f(x)=y$$

So, from (1) we get the differential equation

$$\frac{\text{d} y}{\text{d}x}=\frac{x^{-1}}{\alpha}$$ with the boundary condition $$y(1)=\int_{0}^{\infty} \frac{e^{-u}-e^{-u^{\alpha}}}{u} du =-\frac{(\alpha-1)}{\alpha}\gamma$$

Proof: Take $$y_{alpha}(1)=\psi(\alpha)=\int_{0}^{\infty} \frac{e^{-u}-e^{-u^{\alpha}}}{u} du$$

So, $$\frac{\text{d}\psi(\alpha)}{\text{d}x}=\int_{0}^{\infty} \frac{\partial}{\partial \alpha}\left(\frac{e^{-u}-e^{-u^{\alpha}}}{u}\right) du$$

$$=\int_{0}^{\infty} u^{\alpha-1}\text{ln}(u)e^{-u^{\alpha}} du$$

Taking $$u^{\alpha}=z$$

$$=\frac{1}{{\alpha}^2}\int_{0}^{\infty} \text{ln}(z)e^{-z} dz=-\frac{\gamma}{{\alpha}^2}$$

As $$\Gamma'(1)=-\gamma$$ (https://en.m.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant)

So, we get the differential equation $$\frac{\text{d}\psi}{\text{d}\alpha}=\frac{-\gamma}{{\alpha}^2}$$ with the boundary condition $$\psi(1)=0$$ we get

$$y(1)=\psi(\alpha)-\psi(1)=\psi(\alpha)=-\frac{\alpha-1}{\alpha}\gamma$$

And for, all $$\alpha\geq 1$$, $$y(1)=\frac{-(\alpha-1)\gamma}{\alpha}$$.

Where, $$\gamma$$ is Euler- Mascheroni constant.

So, $$f(x)=\frac{\text{ln}x}{\alpha} -\frac{(\alpha-1)\gamma}{\alpha}$$.

So, for $$\alpha=1$$, $$f(x)=\text{ln}x$$.

• I still think your first equation is wrong as $\int_{0}^{\infty}\frac{\partial}{\mathrm{\partial}x}\frac{e^{-u}-e^{-u^{\alpha}x}}{u}~\mathrm{d}u=\int_{0}^{\infty}\frac{-e^{-u^{\alpha}x}}{u}(-u^{\alpha})~\mathrm{d}u=\int_{0}^{\infty}u^{\alpha-1}e^{-u^{\alpha}x}~\mathrm{d}u$ May 23, 2020 at 13:51
• I see, so you use $\frac{1}{\alpha x}\mathrm{d}z=u^{\alpha-1}~\mathrm{d}u$? May 23, 2020 at 14:22
• How did you get y(1)? Do you have a reference? May 23, 2020 at 14:35
• Isn't that just true for $\alpha =2$? May 23, 2020 at 14:41
• I have added the proof to my solution. May 26, 2020 at 11:45

Numerical Integration vs Analytically Predicted Terms

Error (mostly of Numerical Integration at that magnitude)