28
$\begingroup$

Compute $$\int_0^ex^{1/x}\;\mathrm dx.$$

There is an analytical anti-derivative found in this answer. How does one compute this?

Using the anti-derivative approach we have

$$\int x^{1/x}\;\mathrm d x=x + \frac{\log^2x}{2}-\sum^\infty_{n=2}\sum^n_{k=0}\frac{\log^{n-k}x\;}{x^{n-1}(n-k)!(n-1)^{k+1}}+C$$

Now there is a problem for this anti-derivative as it gets near $0$ (apparent from the $\log$ s). I do not know how to prove this but if $f(x)=x^{1/x}$ and $\lim_{x\to0}f(x)=0$ can be defined then I want to say that $\lim_{x\to0}F(x)=0$. Assuming my non-rigorous logic and some logic from this question we have $$\int_0^ex^{1/x}\;\mathrm dx=\sum _{n=2}^{\infty } \frac{ \Gamma (n+1,\;n-1)}{(n-1)^{n+1}\;n!}+\frac{1}{2}+e$$


Update: Using Mathematica I computed some numerical integrals to find that $F(0)\approx1.53328$. Therefore we have $$\int_0^ex^{1/x}\;\mathrm dx=\sum _{n=2}^{\infty } \frac{ \Gamma (n+1,\;n-1)}{(n-1)^{n+1}\;n!}+\frac{1}{2}+e-\lim_{x\to0}\int x^{1/x}\;\mathrm dx$$

$\endgroup$
2
  • 1
    $\begingroup$ The number is tabulated in oeis.org/A175994 which cites a paper with a graph of the integrad. $\endgroup$
    – user113880
    Dec 6, 2013 at 14:36
  • $\begingroup$ Mathematica 9 gives $2.661825705380417828497039337651395830214970820983303548214678485091470210657175166246828293562435140 \cdots$ $\endgroup$ Jun 17, 2014 at 6:54

3 Answers 3

2
$\begingroup$

Here is a way a physicist is approaching the problem.

First, the variable change $x=\frac{1}{y}$ gives:

$$\int_0^ex^{1/x}\;\mathrm dx=\int_{\frac{1}{e}}^{\infty}\frac{y^{-y}}{y^2}dy$$

Now, as a first approximation, we use the following inequality:

$$\int_{\frac{1}{e}}^{\infty}\frac{y^{-y}}{y^2}dy<\int_{\frac{1}{e}}^{\infty}\frac{y^{-\frac{1}{e}}}{y^2}dy=\int_{\frac{1}{e}}^{\infty}y^{-\frac{1}{e}-2}dy=\frac{e^{2+\frac{1}{e}}}{e+1}$$

Finally:

$$0<\int_0^ex^{1/x}\;\mathrm dx<\frac{e^{2+\frac{1}{e}}}{e+1};\;e>0$$

This result can be probably improved.

$\endgroup$
0
0
$\begingroup$

ISC for 2.66182570538 yields only $$ {\frac {7}{8}}+\sum _{n=1}^{\infty }\frac{60}{ 6\,{5}^{n}+8\,{n}^{3}- 3\,{n}^{2}+7\,n } $$ which is perhaps simpler than your double sum.

$\endgroup$
2
  • $\begingroup$ Thank you for the answer but I am trying to compute this without numerical approximation. Do you reckon that the answer can be found analytically? $\endgroup$ Nov 10, 2013 at 18:05
  • 6
    $\begingroup$ This unfortunately does not converge to the OP's integral. The OP's converges to $$2.6618257053804178284970...$$ while this converges to $$2.661825705395275140900...$$ They disagree on the last digit of your ISC search (and it's odd and unfortunate that your summation, in fact, disagrees with ISC's...) $\endgroup$
    – user98602
    Jan 6, 2014 at 1:09
-1
$\begingroup$

The answer is approximately 2.66183. I hope this helps. I evaluated it, but I am new to the website and wasn't sure how to write out the calculations.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.