Krdytkyu

While thinking about this recent question I thought about an attempt utilizing derivatives of the Beta Function. Sadly I realized that it does not work out for the linked integral but possible for a quite similiar one. To be precise consider the integral

$$mathfrak{I}~=~int_0^{pi}cos(x)^{sin(x)}dx$$

Splitting the integral up in the two intervals $[0,pi/2]$ and $[pi/2,pi]$ and applying the substitution $x=x-pi/2$ to the second one leads to

$$begin{align}
mathfrak{I}=int_0^{pi}cos(x)^{sin(x)}dx&=int_0^{pi/2}cos(x)^{sin(x)}dx+int_{pi/2}^{pi}cos(x)^{sin(x)}dx\
&=int_0^{pi/2}cos(x)^{sin(x)}dx+int_0^{pi/2}sin(x)^{-cos(x)}dx
end{align}$$

Using the fundamental property of the exponential and expanding the integrand as a series yields to

$$begin{align}
mathfrak{I}&=int_0^{pi/2}cos(x)^{sin(x)}dx+int_0^{pi/2}sin(x)^{-cos(x)}dx\
&=int_0^{pi/2}e^{sin(x)log(cos(x))}dx+int_0^{pi/2}e^{-cos(x)log(sin(x))}dx\
&=int_0^{pi/2}sum_{n=0}^{infty}frac{sin^n(x)log^n(cos(x))}{n!}dx+int_0^{pi/2}sum_{n=0}^{infty}frac{(-cos)^n(x)log^n(sin(x))}{n!}dx\
&=sum_{n=0}^{infty}frac1{n!}int_0^{pi/2}sin^n(x)log^n(cos(x))dx+sum_{n=0}^{infty}frac{(-1)^n}{n!}int_0^{pi/2}cos^n(x)log^n(sin(x))dx
end{align}$$

Hence we are dealing with positive integer $n$ we can interpret the inner integral as derivatives of the Beta Function in polar form. Thus the two sums become

$$begin{align}
mathfrak{I}&=sum_{n=0}^{infty}frac1{n!}int_0^{pi/2}sin^n(x)log^n(cos(x))dx+sum_{n=0}^{infty}frac{(-1)^n}{n!}int_0^{pi/2}cos^n(x)log^n(sin(x))dx\
&=sum_{n=0}^{infty}frac1{n!}frac1{2^{n+1}}left.frac{d^n}{dy^n}Bleft(frac{n+1}2,yright)right|_{y=1/2}+sum_{n=0}^{infty}frac{(-1)^n}{n!}frac1{2^{n+1}}left.frac{d^n}{dx^n}Bleft(x,frac{n+1}2right)right|_{x=1/2}
end{align}$$

By using the symmetry of the arguments of the Beta Function we can further simplify this to

$$begin{align}
mathfrak{I}&=sum_{n=0}^{infty}frac1{n!}frac1{2^{n+1}}left.frac{d^n}{dy^n}Bleft(frac{n+1}2,yright)right|_{y=1/2}+sum_{n=0}^{infty}frac{(-1)^n}{n!}frac1{2^{n+1}}left.frac{d^n}{dx^n}Bleft(x,frac{n+1}2right)right|_{x=1/2}\
&=sum_{n=0}^{infty}frac1{n!}frac1{2^{n+1}}left.frac{d^n}{dx^n}Bleft(x,frac{n+1}2right)right|_{x=1/2}+sum_{n=0}^{infty}frac{(-1)^n}{n!}frac1{2^{n+1}}left.frac{d^n}{dx^n}Bleft(x,frac{n+1}2right)right|_{x=1/2}\
&=sum_{n=0}^{infty}frac1{(2n)!}frac1{2^{2n}}left.frac{d^{2n}}{dx^{2n}}Bleft(x,n+frac12right)right|_{x=1/2}
end{align}$$

Presumed all steps done are valid we may conclude that

$$mathfrak{I}~=~int_0^{pi}cos(x)^{sin(x)}dx~=~sum_{n=0}^{infty}frac1{(2n)!}frac1{2^{2n}}left.frac{d^{2n}}{dx^{2n}}Bleft(x,n+frac12right)right|_{x=1/2}$$

Howsoever I have no clue how to proceed from hereon. Hence the derivatives of the Beta Function are getting more and more complicated as $n$ increases, and I am not aware of a simple formula for them, I do not know how to further simplify this monstrosity.

According to WolframAlpha the integral is complex valued for the interval $[0,pi]$ and is approximately $mathfrak{I}=1.03698346829659... + i0.67952701272618...~$. Further note that for the interval $[0,pi/2]$ the value is given by $mathfrak{I}_1=1.05917952027...$ and for $[pi/2,pi]$ given by $mathfrak{I}_2=-0.022196051969... + i0.679527012726...$ which seems to at least justify the separating of the integral numerically.

I am interested in three things. $(1)$ First of all the explicit evaluation of either the integral $mathfrak{I}$ itself or of the derived sum $-$ under the preassumption that this sum in fact equals the integral $-$ if it is in fact possible.$(2)$ Following the latter is my process of approaching valid; if not where did I went wrong?$(3)$ Additionally I want to know how to approach to this kind of sum. For this purpose I would also be satisfied with a evaluation of $mathfrak{I}_1$ or $mathfrak{I}_2$ alone.

Thanks in advance!