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I don't know how to solve these integrals: $$I_1 =\int e^{\Gamma(x)} dx $$ $$I_2 =\int \pi^{\Gamma(x)} dx $$

As a tenth grader I have no idea what the solutions could be. How would one go about evaluating this without computational engines? I'm asking this here because many complex problems have been tackled here...(eg:Integral $\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right) \ \mathrm dx$).

Any hints or solutions to these integrals would be greatly appreciated.

Note: I don't necessarily want closed forms; special functions are okay. [http://en.wikipedia.org/wiki/List_of_mathematical_functions and http://en.wikipedia.org/wiki/Closed-form_expression]

[[ PS: The graphs for the functions inside the aforementioned integrals are amazing! ]]


Background:

I was recently in the process of understanding the wonders of the gamma function. It is really fun to attend to derivatives involving subfactorials, factorials and the gamma function. [In case someone is interested, here are some examples of expressions I was solving] :-

$$ \frac{d}{dx} [x!^{!x}!x^{x!}]^{(x!)(!x)} $$ $$ \frac{d}{dx} \frac{\sqrt{1+\arctan(x)}}{\Gamma(x)} $$

The problem arose when I thought of the two aforementioned integrals I have no answer to.

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    $\begingroup$ @ADG I believe the question is, how does one approach evaluating the integrals $I_1, I_2$ manually. I doubt it is possible to do this in anything like a closed form. $\endgroup$ Feb 24, 2015 at 6:10
  • $\begingroup$ @Travis That's what I thought, but some people on this website have done miracles... So there might be possible solutions.. $\endgroup$ Feb 24, 2015 at 6:12
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    $\begingroup$ It may help a little to note that: $\pi^{f(x)} = (e^{\log \pi})^{f(x)} = e^{\log \pi f(x)}$. Once you've found a closed-form solution the first integral (assumnig there is one), you could probably use this fact to easily find a solution to the second. $\endgroup$
    – Myridium
    Feb 25, 2015 at 13:15
  • $\begingroup$ By 'log' here, do you mean natural log? If so, then yes, that would be useful indeed. $\endgroup$ Feb 25, 2015 at 13:17
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    $\begingroup$ Would the downvoter mind explaining? $\endgroup$ Mar 2, 2015 at 1:21

2 Answers 2

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For these kind of whatever integral you can use a telscoping method explained in

https://math.stackexchange.com/a/4186998

Basically you start from the opposite task

$$\int e^{\Gamma(x)} dx =F(x)$$

$$F'(x) = e^{\Gamma(x)}$$

$$F(x)=e^{\Gamma(x)}f(x)$$

This is a general form we expect

$$F'(x)=e^{\Gamma(x)}=e^{\Gamma(x)}f'(x)+e^{\Gamma(x)}\Gamma(x)\Gamma'(x)f(x)$$

This is making

$$f(x)=\frac{1}{\Gamma(x)\Gamma'(x)}$$

the best guess. The rest is simply recursive

$$\int e^{\Gamma(x)} dx = e^{\Gamma(x)} \sum_{n=0}^{+\infty}f_n(x)$$

$$f_0(x)=\frac{1}{\Gamma(x)\Gamma'(x)}$$

$$f_{n+1}(x)=-\frac{f_{n}'(x)}{\Gamma'(x)}$$

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  • $\begingroup$ @TymaGaidash I have no accounts. It is just StackExchange combining answers. $\endgroup$
    – user953399
    Jul 27, 2021 at 6:29
  • $\begingroup$ @TymaGaidash It is a generalized method used in particular here and there, for example for asymptotic series of Dawson's integral and so on. This is just the essence of it applicable in general for much wider class of integrals. $\endgroup$
    – user953399
    Jul 27, 2021 at 8:00
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These will simply be many small solutions. Here is the graph of the function. Note that I will take the primitive for simplicity.

The first way uses the Maclaurin series for $e^y$. We can integrate term by term:

$$\mathrm{A(a)=\int a^{Γ(x)}dx=\int \sum_{n=0}^\infty \frac{ln^n(a)Γ^n(x)}{n!}dx= \sum_{n=0}^\infty \frac{ln^n(a)}{n!}\int Γ^n(x)dx= \quad\sum_{n=0}^\infty \frac{ln^n(a)}{n!}\int \prod_{n=1}^{x-1} (x-m)^ndx}$$

It would be nice if we could use a truncated version of the Pentagonal Number theorem or we could expand the $(x-m)^n$ into a binomial series. We can also turn the product into a sum via logarithms. I still do not see how to integrate this.

Here is another idea which usually works for integration. Integrating with a Taylor series or using the Riemann Sum definition:

$$\mathrm{\int_a^b y^{Γ(x)}dx=\lim_{n\to\infty}\frac{b-a}{n}\sum_{k=0}^n y^{Γ\left(a+k\frac{b-a}{n}\right)}=\sum_{n=0}^\infty \frac{\frac{d^n}{dx^n}\left(y^{Γ(x)}\right )\big|_{x=r}(x-r)^n}{n!}}$$

I also had the idea to expand $Γ^n(x)$ into factored integral representations of Γ(x) and using the Cauchy repeated integral formula. I tried it, but am unsure if it would be used correctly.

I will add more. Please correct me and give me feedback!

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