While I was playing with Wolfram Alpha online calculator I wondered how calculate an approximation of $$I=\int_0^1 \log\left(1+x^{\Gamma(x)}\right)\,dx,\tag{1}$$ where $\Gamma(x)$ is the Gamma function. I am not interested specially to get a very good approximation, because Wolfram Alpha provide me one with code

int log(1+x^Gamma(x))dx, from x=0 to 1

I would like to know methods about how calculate an approximation of $I$. I suspect that to get a good approximation is required to do analysis about the Gamma function.

Question. How calculate using analysis an approximation of $$\int_0^1 \log\left(1+x^{\Gamma(x)}\right)\,dx?$$ Many thanks.

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    $\begingroup$ Any reliable numerical integrator should do it, e.g. en.wikipedia.org/wiki/QUADPACK $\endgroup$ – gammatester Sep 3 '17 at 19:12
  • $\begingroup$ Many thanks @gammatester that I would like to know is how do enough good calculations using for example analysis/calculus to get an approximation. Any case many thanks for your help. $\endgroup$ – user243301 Sep 3 '17 at 19:16
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    $\begingroup$ There's no analysis needed, you have an endpoint singularity at $x=0$, but it's a harmless one. Just use an IMT-type rule. $\endgroup$ – Professor Vector Sep 3 '17 at 20:01
  • $\begingroup$ Many thanks @ProfessorVector I am going to wait if some user has a different approach, but in any case thanks for your help. $\endgroup$ – user243301 Sep 3 '17 at 20:09
  • $\begingroup$ Sorry for the insistence @ProfessorVector and do you think that this (different) question could be interesting: study how calculate a good approximation of integrals of the form $\int_0^1 \log\left(1+x^{f(x)}\right)\,dx$, where $f(x)$ is a known function. Also use an IMT-type rule solve this new question, without analysis? Many thanks. $\endgroup$ – user243301 Sep 4 '17 at 15:32

First we note that within $(0,1)$ we have $x^{\Gamma(x)}<x$.

From this we get immediately an upper bound for $I$:

$$\int_0^1 \log\left(1+x\right)\,dx=2\ln2-1=0.3862...$$

If we are not satisfied with this robust estimation then within $(0,1)$ a good approximation for $\Gamma(x)$ is $\frac {1}{x}$

So we get a better estimation for $I$:

$$\int_0^1 \log\left(1+x^{\frac {1}{x}}\right)\,dx$$

Since $0<x^{\frac {1}{x}}<1$ we can use Taylor series for the logarithm

$$\int_0^1 \log\left(1+x^{\frac {1}{x}}\right)\,dx=f(1)-\frac {f(2)}{2}+...+\frac {(-1)^{(k+1)}f(k)}{k}+...$$


$$f(k)=\int_0^1 x^{\frac {k}{x}}d\,x$$

For example $\frac {f(10)}{10}=0.0079...$ but since we calculate only an estimation i think it is needed fewer terms to get a satisfactory estimate.

  • $\begingroup$ Many thanks to you, to user90369 and Professor Vector. I know that some times my questions weren't the best, but my intention is learn and propose new exercises. Many thanks for your answer I am going to study it. $\endgroup$ – user243301 Sep 10 '17 at 10:02

Only a note.

You can use Riemann sums (e.g. https://en.wikipedia.org/wiki/Riemann_sum) for a first approximation.

Be $\enspace\displaystyle f(n):=\sqrt[n]{\prod\limits_{k=1}^n(1+(\frac{k}{n})^{\Gamma(\frac{k}{n})})} \,$ . Then we get $\enspace\displaystyle \int\limits_0^1 \ln(1+x^{\Gamma(x)})dx=\ln\lim\limits_{n\to\infty}f(n)\,$ .

It's $\enspace\displaystyle \int\limits_0^1 \ln(1+x^{\Gamma(x)})dx =0,2831535412286936...$

and e.g. $\enspace\ln f(10000)\approx 0.283188199004388115\,$ .

Disadvantage: The convergence is very slow.

  • $\begingroup$ Many thanks for your courage and contribution. I am going to study your answer, and I hope more contributions. $\endgroup$ – user243301 Sep 4 '17 at 21:00
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    $\begingroup$ @user243301 : It was a pleasure. :-) But: It's needed something which convergies fast. $\endgroup$ – user90369 Sep 5 '17 at 5:11

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