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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?

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2 Answers 2

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

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

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

Error (mostly of Numerical Integration at that magnitude) enter image description here

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