反三角函数积分级数证明

╰☆ヾo.海x

For $a\ll1$ show that$$\int_{0}^{1}2\arctan(1+a\ln(x))dx=\frac{\pi}{2}-a-a^2-a^3+O(a^4)$$

战巡

回复 1# ╰☆ヾo.海x


\[2\arctan(1+a\ln(x))=\frac{\pi}{2}+\ln(x)a-\frac{\ln^2(x)}{2}a^2+\frac{\ln^3(x)}{6}a^3+O(a^4)\]
剩下积分自己做吧

╰☆ヾo.海x

回复 2# 战巡

怎么求出来的啊？

战巡

回复 3# ╰☆ヾo.海x

就马克劳林展开啊

hbghlyj

回复 1# ╰☆ヾo.海x

math.stackexchange.com/questions/640220/
Your integral is a function of $a$:
$$I(a)=2\int_0^1 dx \arctan(1+a\ln x).$$
Your strategy to use the Taylor expansion was a good idea, but you should expand in $a$ instead of $x$ and around $0$ instead of $1$, because you need $a\ll 1$. So we seek for the coefficients in the expression
\begin{equation}I(a)=I(0)+I'(0)a+\tfrac{1}{2!}I''(0)a^2+\tfrac{1}{3!}I'''(0)a^3+O(a^4).\label1\end{equation}
We will first differentiate the integrand $f(a,x):=\arctan(1+a\ln x)$ w.r.t $a$:
$$\partial_a f(a,x)=\frac{\ln x}{1+(1+a\ln x)^2}$$
As you say it becomes messy, the trick is here to reexpress the new derivatives using the previous:
$$\partial_a^2f(a,x)=({\text{standard rules}})=\frac{-(\ln x)^2}{(1+(1+a\ln x)^2)^2}2(1+a\ln x)=-2(1+a\ln x)(\partial_a f)^2$$
Now this will save us a lot of work:
$$\partial_a^3f(a,x)=-2\ln x (\partial_a f)^2-2(1+a\ln x)\partial_a(\partial_a f)^2=-2\ln x(\partial_a f)^2-4(1+a\ln x)(\partial_a f)(\partial_a^2 f)$$
Next we just set $a=0$ and see what happens
$$f(0,x)=\arctan(1)=\tfrac{\pi}{4}$$
$$\partial_a f(0,x)=\tfrac{1}{2}\ln x$$
$$\partial_a^2 f(0,x)=-\tfrac{1}{2}(\ln x)^2$$
$$\partial_a^3 f(0,x)=\tfrac{1}{2}(\ln x)^3$$
Now we can compute the coefficients of the Taylor expansion
$$I(0)=2\int_0^1 dx f(0,x)=\tfrac{\pi}{2}$$
$$I'(0)=2\int_0^1 dx \partial_a f(0,x)=\int_0^1 dx \ln x =(x\ln x -x)_{\epsilon\to 0}^1=-1$$
$$I''(0)=2\int_0^1 dx \partial_a^2 f(0,x)=-\int_0^1 dx (\ln x)^2 =-(x(\ln x)^2 -2x\ln x+2x)_{\epsilon\to 0}^1=-2$$
$$I'''(0)=2\int_0^1 dx \partial_a^3 f(0,x)=\int_0^1 dx (\ln x)^3 =(x(\ln x)^3-3x(\ln x)^2+6x\ln x-6x)_{\epsilon\to 0}^1=-6$$
To keep it short, I did not explain how to do $\int dx (\ln x)^n$, please comment if you need further help on that!

Plugging everything into \eqref{1} gives:
$$I(a)=\tfrac{\pi}{2}-a-a^2-a^3+O(a^4)$$