两个奇怪的积分
Evaluate integral
$$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}$$
Well,I think we have
$$\int_{0}^{1}{x^{-x}(1-x)^{x-1}\sin{\pi x}dx}=\frac{\pi}{e}$$
and
$$\int_{0}^{1}{x^{x}(1-x)^{1-x}\sin{\pi x}dx}=\frac{e\pi}{24}$$
With such nice result of these integral,why isn't worth to evaluate it?
I found a solution about the second one,but I wonder it will work for the first one
Note
$$ S=\int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}-\int_{0}^{1}{(1-x)e^{(i\pi+\ln{x}-\ln{(1-x)})x}dx} $$
Let $t=\ln{x}-\ln{(1-x)}$,$x=\frac{e^{t}}{1+e^{t}}$
Thus
\begin{align}S&=\int_{-\infty}^{+\infty}{\frac{1}{e^{t}+1}e^{(i\pi+t)\frac{e^{t}}{1+e^t}}\frac{e^{t}}{(1+e^{t})^{2}}dt}\\ &=\int_{-\infty+i\pi}^{-\infty-i\pi}{e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}}dt}\end{align}
Due to
$$ f(z)=e^{\frac{te^{t}}{e^{t}-1} } \frac{e^{t}}{(e^{t}-1)^{3}},\qquad D=\{Z\in C|-\pi\leq Im(z) \leq \pi\}$$
Therefore
$res(f,0)=-\frac{e}{24}$when $z=0$
with $ \zeta_{R}=\gamma_{R}+o_{R}+\tau_{R}$
$$\oint_{\zeta_{R}}{f(z)dz}=-2\pi i\cdot res(f,0)=\frac{2i\pi e}{24}$$
because
$$ \{z_{n}\}\subset D,\qquad |z_{n}|\rightarrow\infty $$
Therefore
$$ 2S=2\lim_{R\rightarrow \infty}\int_{\gamma_{R}}{f(z)dz} $$
gives
$$ \int_{0}^{1}{\sin{\pi x}x^{x}(1-x)^{1-x}dx}=Im(S)=\frac{e\pi}{24} $$
My friend tian_275461 told me he use a simliar method to deal with the first one to obtain the result $\frac{\pi}{e}$,but I am not figure it out.
第一个积分的解答:
Exactly the same method works for the other case.
$$\int_0^1 x^{-x} (1-x)^{x-1}\sin{\pi x} dx = \mathrm{Im}\left[\int_0^1 \frac{e^{(i\pi+\ln(1-x)-\ln x)x}}{1-x}dx\right]$$
Write $t=\ln((1-x)/x)$ and $z=t+i\pi$ as you did above to get
$$S = \int_0^1 \frac{e^{(i\pi+\ln(1-x)-\ln x)x}}{1-x}dx=\int_{-\infty+i\pi}^{\infty+i\pi} \frac{e^{\frac{z}{1-e^z}}}{1-e^z}dz$$
Then with $$f(z)=\frac{e^{\frac{z}{1-e^z}}}{1-e^z}$$
the only pole is at $z=0$, $res(f,0)=-\frac{1}{e}$ and in the limit $2S = \oint f(z)dz=-2\pi i \cdot res(f,0) = 2\pi i/e$ and your answer follows.
第二个积分的另一种求法:
This one can be done with "residue at infinity" calculation. This method is shown in the Example VI of http://en.wikipedia.org/wiki/Methods_of_contour_integration .
First, we use $z^z = \exp ( z \log z )$ where $\log z$ is defined for $-\pi\leq \arg z < \pi$.
For $(1-z)^{1-z} = \exp ( (1-z)\log (1-z) )$, we use $\log (1-z)$ defined for $0\leq \arg(1-z) <2\pi$.
Then, let $f(z)= \exp( i\pi z + z \log z + (1-z) \log (1-z) )$.
As shown in the Ex VI in the wikipedia link, we can prove that $f$ is continuous on $(-\infty, 0)$ and $(1,\infty)$, so that the cut of $f(z)$ is $[0,1]$.
We use the contour: (consisted of upper segment: slightly above $[0,1]$, lower segment: slightly below $[0,1]$, circle of small radius enclosing $0$, and circle of small radius enclosing $1$, that looks like a dumbbell having knobs at $0$ and $1$, can someone edit this and include a picture of it please? In fact, this is also the same contour as in Ex VI, with different endpoints.)
On the upper segment, the function $f$ gives, for $0\leq r \leq 1$,
$$\exp(i\pi r) r^r (1-r)^{1-r} \exp( (1-r) 2\pi i ).$$
On the lower segment, the function $f$ gives, for $0\leq r \leq 1$,
$$\exp(i\pi r) r^r(1-r)^{1-r}. $$
Since the functions are bounded, the integrals over circles vanishes when the radius tend to zero.
Thus, the integral of $f(z)$ over the contour, is the integral over the upper and lower segments, which contribute to
$$\int_0^1 \exp(i\pi r) r^r (1-r)^{1-r} dr - \int_0^1 \exp(-i\pi r) r^r(1-r)^{1-r} dr$$
which is
$$2i \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr.$$
By the Cauchy residue theorem, the integral over the contour is
$$-2\pi i \textrm{Res}_{z=\infty} f(z) = 2\pi i \textrm{Res}_{z=0} \frac{1}{z^2} f(\frac 1 z).$$
From a long and tedious calculation of residue, it turns out that the value on the right is
$$2i \frac{\pi e}{24}.$$
Then we have the result:
$$ \int_0^1 \sin(\pi r) r^r (1-r)^{1-r} dr = \frac{\pi e}{24}.$$
我们也可得到\begin{align*} \int_{0}^{1} e^{i \pi x} \, x^{x} (1-x)^{1-x} \, dx = i \, \frac{\pi e}{4!} \end{align*}
来自:http://math.stackexchange.com/questions/324647/integrate-int-01x-x1-xx-1-sin-pi-xdx
http://math.stackexchange.com/questions/958624/prove-that-int-01-sin-pi-xxx1-x1-x-dx-frac-pi-e24
2022年8月20日 01:31
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