What is the shape of $\int_0^1 P_n(x)\pi^xdx$? Using wolfram, to evaluate a few values of
$$\int_0^1x^n\pi^xdx$$
I think it's has the shape $(\pi p_n(\ln\pi)+(-1)^nn!)/\ln^{n+1}\pi$, where $p_n(\ln\pi)$ is a polynomial in $\ln\pi$ with degree $n$ and integer coefficients. Can someone say whats the shape of the integral when we change $x^n$ for a polynomial with degree $n$ and integer coeffcients?
An example from wolfram is:
$\int(6x^2-6x+1)\pi^xdx= (π^x (12 + 6 \log(π) + \log^2(π) + 6 x^2 \log^2(π) - 6 x \log(π) (2 + \log(π))))/(\log^3(π))$
from $0$ to $1$ is $(-12 + 12 π - 6 \log(π) - 6 π \log(π) - \log^2(π) + π \log^2(π))/(\log^3(π))$
which seems to be a polynomial in $\ln\pi$ with degree $n$ multiplied by $\pi$ plus another polynomial in $\ln\pi$ with degree $n$ with all divided by $\ln^{n+1}\pi$ . So I guess it has the shape $[\pi p_n(\ln\pi)+q_n(\ln\pi)]/\ln^{n+1}\pi$.
 A: For any real $b>0$,
$$\int_0^1 x^n b^x \, dx = \int_0^1 x^n e^{\ln(b)\,x} \, dx = \frac1{\ln^{n+1}(b)} \int_0^{\ln(b)} x^n e^x \, dx$$
It's easy to find a recurrence relation for the remaining integral by parts.
Then upon swapping out $x^n$ for the polynomial $P_n(x) = \sum\limits_{i=0}^n a_ix^i$, we have
$$\int_0^1 P_n(x) b^x \, dx = \sum_{i=0}^n \frac{a_i}{\ln^{i+1}(b)} \int_0^{\ln(b)} x^i e^x \, dx$$
A: Well, a general polynomial can be expressed as follows:
$$\text{y}_\text{n}\left(x\right):=\sum_{\text{k}\space=\space0}^\text{n}\alpha_\text{k}x^\text{k}\tag1$$
So, we get:
$$\mathcal{I}_\text{n}\left(\beta\right):=\int_0^1\text{y}_\text{n}\left(x\right)\beta^x\space\text{d}x=\int_0^1\sum_{\text{k}\space=\space0}^\text{n}\alpha_\text{k}x^\text{k}\beta^x\space\text{d}x=\sum_{\text{k}\space=\space0}^\text{n}\alpha_\text{k}\int_0^1x^\text{k}\beta^x\space\text{d}x\tag2$$
If, we substitute $\text{u}=x^{\text{k}+1}$ we get:
$$\mathcal{I}_\text{n}\left(\beta\right)=\sum_{\text{k}\space=\space0}^\text{n}\frac{\alpha_\text{k}}{\text{k}+1}\int_0^1\beta^{\text{u}^\frac{1}{\text{k}+1}}\space\text{du}\tag3$$
And this integral is defined using the incomplete gamma function:
$$\int\beta^{\text{u}^\frac{1}{\text{k}+1}}\space\text{du}=\frac{\text{k}+1}{\left(-1\right)^{\text{k}+2}}\cdot\frac{1}{\ln^{\text{k}+1}\left(\beta\right)}\cdot\Gamma\left(\text{k}+1,-\ln\left(\beta\right)\text{u}^\frac{1}{\text{k}+1}\right)\tag4$$
