Evaluate the definite integral $\int _0^1x^bb^x\,dx$. Since, the question asks to evaluate  $\displaystyle \int \limits _0^1x^bb^x\,dx$. Sharing my thought shots on the same:
I multiplied and divided my integrand with $\log b^x$ and afterwards substituted $b^x=t$. So my Integrand reduced to $\displaystyle \int \limits _1^b\frac{(\log t)^b}{(\log b)^{b+1}}\,dt$. I can't see things getting better ahead. If someone can help!
Thanks in advance.
Please note that this is not the case with $b\to \infty$.
 A: This integral looks like a gamma function:
$$\int_0^1 b^x x^b dx=\int_0^1 e^{\ln(b)x}x^b dx$$
Take $-\frac y{\ln(b)}=x$:
$$\int_0^{-\ln(b)} e^{-y} \left(-\frac y{\ln(b)}\right)^b\cdot -dy\frac1{\ln(b)}=(-\ln(b))^{-b-1}\int_0^{-\ln(b)} e^{-y} y^b dy$$
Remember the Incomplete Gamma function $\gamma(a,x)=\int_0^x k^{a-1} e^{-k}dk$:
$$(-\ln(b))^{-b-1}\int_0^{-\ln(b)} e^{-y} y^b dy =\frac{γ(b+1,-\ln(b))}{(-\ln(b))^{b+1}}$$
Here is a tester of the result which works for all $b>0$, but also works for some $b\in\Bbb C$
A: After getting the integral as:
\begin{align}\int_{1}^{b}\frac{(\ln t)^b}{(\ln b)^{b+1}}\,dx\end{align}
It can be seen that upon using the substitution $\ln t = \xi$, the integral gets transformed into:
\begin{align}\frac{1}{(\ln b)^{b+1}}\int_{0}^{\ln b}\xi^be^{\xi}\,d\xi\end{align}
If b is a positive integer, then a recursive method is applicable. If b is a number which is not an integer, we transform the integral
\begin{align}\frac{1}{(\ln b)^{b+1}}\int_{0}^{-\ln \frac{1}{b}}\xi^be^{\xi}\,d\xi=-\frac{(-1)^b}{(\ln b)^{b+1}}\int_{0}^{\ln \frac{1}{b}}\xi^be^{-\xi}\,d\xi\end{align}
\begin{align}=\frac{(-1)^{b+1}}{(\ln b)^{b+1}}\int_{0}^{\ln \frac{1}{b}}\xi^be^{-\xi}\,d\xi=\frac{1}{(\ln \frac{1}{b})^{b+1}}\int_{0}^{\ln \frac{1}{b}}\xi^be^{-\xi}\,d\xi\end{align}
Further, the Incomplete Gamma Function can be used to give
$$
(\ln (\frac{1}{b}))^{-b-1}\gamma(b+1,\ln (\frac{1}{b}))
$$
Coming to the recursion formula, suppose $f(x)=\xi^b$ and $g(x)=e^{\xi}$. Then we know that $g'(x)=g(x)$
This answer explains the recursion, so the solution is
\begin{align}
 \frac{1}{(\ln b)^{b+1}}\bigg[\sum\limits_{k = 0}^b {( - 1)^{b - k} \frac{{b!}}{{k!}}\xi^k } \bigg]e^{\xi}\Biggr|_{0}^{\ln b}
\end{align}
