请问这两个积分怎么算？

dim

谢谢啦

dim

答案是这两个，求大神支招

战巡

回复 1# dim

我只提供方法，其中计算过于繁琐，你自己看着办
令
\[f(a,x)=\frac{\arctan(ax^4)}{1+x^2}\]
\[\frac{\partial}{\partial a}f(a,x)=\frac{x^4}{(1+x^2)(1+a^2x^8)}=\frac{1}{(1+a^2)(1+x^2)}+\frac{-a^2x^6+a^2x^4+x^2-1}{(1+a^2)(1+a^2x^8)}\]
\[\int_0^{+\infty}f(2,x)dx=\int_0^{+\infty}\int_0^2\frac{\partial}{\partial a}f(a,x)dadx+\int_0^{+\infty}f(0,x)dx=\int_0^2da\int_0^{+\infty}\frac{\partial}{\partial a}f(a,x)dx+\frac{\pi}{2}\]
接下来就是最讨厌的部分了，你要强行算出
\[\int_0^{+\infty}\frac{\partial}{\partial a}f(a,x)dx=\int_0^{+\infty}[\frac{1}{(1+a^2)(1+x^2)}+\frac{-a^2x^6+a^2x^4+x^2-1}{(1+a^2)(1+a^2x^8)}]dx\]
前半部分好弄，后面你只能按有理分式积分的那套，强行拆解为如下形式
\[\frac{-a^2x^6+a^2x^4+x^2-1}{(1+a^2)(1+a^2x^8)}=\frac{A_1+B_1x}{x^2+b_1x+c_1}+\frac{A_2+B_2x}{x^2+b_2x+c_2}+\frac{A_3+B_3x}{x^2+b_3x+c_3}+\frac{A_4+B_4x}{x^2+b_4x+c_4}\]
而后逐个积分，具体我不写了，直接出结果
\[\int_0^{+\infty}\frac{\partial}{\partial a}f(a,x)dx=\frac{\pi}{2}·\frac{1}{1+a^2}+\frac{\pi(1+\sqrt{a})(\sqrt{4-2\sqrt{2}}a-2\sqrt{2+\sqrt{2}}\sqrt{a}+\sqrt{4-2\sqrt{2}})}{8a^{\frac{3}{4}}(1+a^2)}\]
再对这个求积分，换元$y=a^{\frac{1}{4}}$后又变成类似上面的有理分式，再做一次，最后出结果

第二个也差不多，反正这样都能做下去

注：不要问我什么有没有简单一点的方法，我懒得再管这个问题

dim

回复 3# 战巡


    好的，谢谢大神！我还想问一下这是你手算出来的吗？还是用了什么软件？

caijinzhi

转化为含参变量的积分

青青子衿

回复 3# 战巡
\begin{align*}
f(a,x)&=\dfrac{\arctan\left(a\,x^4\right)}{1+x^2}\\
F(a,x)&=\int_0^{+\infty}\dfrac{\arctan\left(a\,x^4\right)}{1+x^2}\mathrm{d}x=\int_0^{+\infty}f(a,x)\mathrm{d}x\\
\dfrac{\partial}{\partial\,\!a}f(a,x)&=\dfrac{x^4}{\left(1+x^2\right)\left(1+a^2x^8\right)}\\
&=\dfrac{1}{\left(1+a^2\right)\left(1+x^2\right)}-\dfrac{a^2x^6-a^2x^4-x+1}{\left(1+a^2\right)\left(1+a^2x^8\right)}\\
&=\dfrac{1}{\left(1+a^2\right)\left(1+x^2\right)}-\dfrac{\left(x+1\right)\left(x-1\right)\left(a\,\!x^2+1\right)\left(a\,\!x^2-1\right)}{\left(1+a^2\right)\left(1+a^2x^8\right)}\\
f(a,x)&=\dfrac{\arctan\left(a\,x^4\right)}{1+x^2}=\int_0^a\dfrac{x^4}{\left(1+x^2\right)\left(1+t^2x^8\right)}\mathrm{d}t\\
&=\int_0^a\dfrac{1}{\left(1+x^2\right)\left(1+t^2\right)}\mathrm{d}t-\int_0^a\dfrac{t^2x^6-t^2x^4-x^2+1}{\left(1+t^2\right)\left(1+t^2x^8\right)}\mathrm{d}t\\
\end{align*}
\begin{align*}
F(a,x)&=\int_0^{+\infty}\dfrac{\arctan\left(a\,x^4\right)}{1+x^2}\mathrm{d}x=\int_0^{+\infty}f(a,x)\mathrm{d}x\\
&=\int_0^{+\infty}\mathrm{d}x\int_0^a\dfrac{x^4}{\left(1+x^2\right)\left(1+t^2x^8\right)}\mathrm{d}t=\int_0^a\mathrm{d}t\int_0^{+\infty}\dfrac{x^4}{\left(1+x^2\right)\left(1+t^2x^8\right)}\mathrm{d}x\\
&=\int_0^a\mathrm{d}t\int_0^{+\infty}\dfrac{1}{\left(1+x^2\right)\left(1+t^2\right)}\mathrm{d}x-\int_0^a\mathrm{d}t\int_0^{+\infty}\dfrac{t^2x^6-t^2x^4-x^2+1}{\left(1+t^2\right)\left(1+t^2x^8\right)}\mathrm{d}x\\
&=\int_0^a\dfrac{1}{1+t^2}\mathrm{d}t\int_0^{+\infty}\dfrac{1}{1+x^2}\mathrm{d}x-\int_0^a\mathrm{d}t\int_0^{+\infty}\dfrac{\sqrt{\,t\,}\,u^6-t\,u^4-\frac{1}{\sqrt{\,t\,}}u^2+1}{\left(1+t^2\right)\left(1+u^8\right)\sqrt[4]{\,t\,}}\mathrm{d}u\\
&=\dfrac{\pi}{2}\arctan\left(a\right)+\dfrac{\sqrt{2\sqrt{2}\left(\sqrt{2}+1\right)}}{8}\pi\int_0^a\dfrac{\left(1+\sqrt{\,t\,}\,\right)\bigg(\left(\sqrt{2}-1\right)-\sqrt{2\,t\,}\,+\left(\sqrt{2}-1\right)t\bigg)}{\left(1+t^2\right)\sqrt[4]{\,t^3\,}}\mathrm{d}t\\
&=\dfrac{\pi}{2}\arctan\left(a\right)+\dfrac{\pi}{8}\csc\left(\dfrac{\pi}{8}\right)\int_0^{\sqrt[4]{a}}\dfrac{\left(1+v^2\right)\bigg(\left(\sqrt{2}-1\right)-\sqrt{2}\,v^2+\left(\sqrt{2}-1\right)v^4\bigg)}{\left(1+v^8\right)v^3}\left(4v^3\right)\mathrm{d}v\\
&=\dfrac{\pi}{2}\arctan\left(a\right)+\dfrac{\pi}{2}\csc\left(\dfrac{\pi}{8}\right)\int_0^{\sqrt[4]{a}}\dfrac{\left(1+v^2\right)\bigg(\left(\sqrt{2}-1\right)-\sqrt{2}\,v^2+\left(\sqrt{2}-1\right)v^4\bigg)}{1+v^8}\mathrm{d}v\\
\end{align*}
\begin{align*}
F(a)
&=\dfrac{\pi}{2}\arctan\left(a\right)+\dfrac{\pi}{2}\csc\left(\dfrac{\pi}{8}\right)\int_0^{\sqrt[4]{a}}\dfrac{\left(1+v^2\right)\bigg(\left(\sqrt{2}-1\right)-\sqrt{2}\,v^2+\left(\sqrt{2}-1\right)v^4\bigg)}{1+v^8}\mathrm{d}v\\
\\
&=\qquad
\begin{split}
\dfrac{\pi}{2}\arctan\left(a\right)\qquad\qquad\qquad\qquad\quad\,\,\,\,\\
-\dfrac{\pi}{2}\arctan\left(\sqrt[4]{a}\csc\left(\dfrac{\pi}{8}\right)-\cot\left(\dfrac{\pi}{8}\right)\right)\\
-\dfrac{\pi}{2}\arctan\left(\sqrt[4]{a}\csc\left(\dfrac{\pi}{8}\right)+\cot\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{2}\arctan\left(\sqrt[4]{a}\sec\left(\dfrac{\pi}{8}\right)-\tan\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{2}\arctan\left(\sqrt[4]{a}\sec\left(\dfrac{\pi}{8}\right)+\tan\left(\dfrac{\pi}{8}\right)\right)
\end{split}
\\
\end{align*}
...

1. Integrate[ArcTan[2*x^4]/(1 + x^2), {x, 0, +Infinity}] // N
2. (Pi/2*ArcTan[2]
3.    - Pi/2*ArcTan[Power[2, 1/4]*Csc[Pi/8] - Cot[Pi/8]]
4.    - Pi/2*ArcTan[Power[2, 1/4]*Csc[Pi/8] + Cot[Pi/8]]
5.    + Pi/2*ArcTan[Power[2, 1/4]*Sec[Pi/8] - Tan[Pi/8]]
6.    + Pi/2*ArcTan[Power[2, 1/4]*Sec[Pi/8] + Tan[Pi/8]]) // N

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青青子衿

回复 3# 战巡

\begin{align*}
g(a,x)&=\dfrac{\arctan\left(a\,x^4\right)}{1+x^4}\\
G(a)&=\int_0^{+\infty}\dfrac{\arctan\left(a\,x^4\right)}{1+x^4}\mathrm{d}x=\int_0^{+\infty}g(a,x)\mathrm{d}x\\
\dfrac{\partial}{\partial\,\!a}g(a,x)&=\dfrac{\partial}{\partial\,\!a}\dfrac{\arctan\left(a\,x^4\right)}{1+x^4}=\dfrac{x^4}{\left(1+x^4\right)\left(1+a^2x^8\right)}\\
&=\dfrac{a^2x^4+1}{\left(1+a^2\right)\left(1+a^2x^8\right)}-\dfrac{1}{\left(1+a^2\right)\left(1+x^4\right)}\\
g(a,x)&=\dfrac{\arctan\left(a\,x^4\right)}{1+x^4}=\int_0^a\dfrac{x^4}{\left(1+x^4\right)\left(1+t^2x^8\right)}\mathrm{d}t\\
&=\int_0^a\dfrac{t^2x^4+1}{\left(1+t^2\right)\left(1+t^2x^8\right)}\mathrm{d}t-\int_0^a\dfrac{1}{\left(1+x^4\right)\left(1+t^2\right)}\mathrm{d}t\\
\end{align*}
\begin{align*}
G(a)&=\int_0^{+\infty}\dfrac{\arctan\left(a\,x^4\right)}{1+x^4}\mathrm{d}x=\int_0^{+\infty}g(a,x)\mathrm{d}x\\
&=\int_0^{+\infty}\mathrm{d}x\int_0^a\dfrac{x^4}{\left(1+x^4\right)\left(1+t^2x^8\right)}\mathrm{d}t=\int_0^a\mathrm{d}t\int_0^{+\infty}\dfrac{x^4}{\left(1+x^4\right)\left(1+t^2x^8\right)}\mathrm{d}x\\
&=\int_0^a\mathrm{d}t\int_0^{+\infty}\dfrac{t^2x^4+1}{\left(1+t^2\right)\left(1+t^2x^8\right)}\mathrm{d}x-\int_0^a\mathrm{d}t\int_0^{+\infty}\dfrac{1}{\left(1+x^4\right)\left(1+t^2\right)}\mathrm{d}x\\
&=\int_0^a\mathrm{d}t\int_0^{+\infty}\dfrac{t\,u^4+1}{\left(1+t^2\right)\left(1+u^8\right)\sqrt[4]{\,t\,}}\mathrm{d}u-\int_0^a\dfrac{1}{1+t^2}\mathrm{d}t\int_0^{+\infty}\dfrac{1}{1+x^4}\mathrm{d}x\\
&=\dfrac{\pi}{8}\int_0^a\dfrac{\,t\cdot\sqrt{2\sqrt{2}\left(\sqrt{2}-1\right)}+\sqrt{2\sqrt{2}\left(\sqrt{2}+1\right)}}{\left(1+t^2\right)\sqrt[4]{\,t\,}}\mathrm{d}t-\dfrac{\pi}{2\sqrt{2}}\arctan\left(a\right)\\
&=\dfrac{\pi}{8}\int_0^a\dfrac{\,t\,\sec\left(\dfrac{\pi}{8}\right)+\csc\left(\dfrac{\pi}{8}\right)}{\left(1+t^2\right)\sqrt[4]{\,t\,}}\mathrm{d}t-\dfrac{\pi}{2\sqrt{2}}\arctan\left(a\right)\\
&=\dfrac{\pi}{8}\int_0^{\sqrt[4]a}\dfrac{\,v^4\,\sec\left(\dfrac{\pi}{8}\right)+\csc\left(\dfrac{\pi}{8}\right)}{\left(1+v^8\right)v}\left(4v^3\right)\mathrm{d}v-\dfrac{\pi}{2\sqrt{2}}\arctan\left(a\right)\\
&=\dfrac{\pi}{2}\int_0^{\sqrt[4]a}\dfrac{\,v^4\,\sec\left(\dfrac{\pi}{8}\right)+\csc\left(\dfrac{\pi}{8}\right)}{1+v^8}\left(v^2\right)\mathrm{d}v-\dfrac{\pi}{2\sqrt{2}}\arctan\left(a\right)
\end{align*}
\begin{align*}
G(a)
&=\dfrac{\pi}{2}\int_0^{\sqrt[4]a}\dfrac{\,v^4\,\sec\left(\dfrac{\pi}{8}\right)+\csc\left(\dfrac{\pi}{8}\right)}{1+v^8}\left(v^2\right)\mathrm{d}v-\dfrac{\pi}{2\sqrt{2}}\arctan\left(a\right)\\
\\
&=\qquad
\begin{split}
-\dfrac{\pi}{2\sqrt2}\arctan\left(a\right)\qquad\qquad\qquad\qquad\quad\,\,\,\,\\
+\dfrac{\pi}{2\sqrt2}\arctan\left(\sqrt[4]{a}\csc\left(\dfrac{\pi}{8}\right)-\cot\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{2\sqrt2}\arctan\left(\sqrt[4]{a}\csc\left(\dfrac{\pi}{8}\right)+\cot\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{4\sqrt2}\ln\left(1-2\sqrt[4]{a}\sin\left(\dfrac{\pi}{8}\right)+\sqrt{a}\right)\\
-\dfrac{\pi}{4\sqrt2}\ln\left(1+2\sqrt[4]{a}\sin\left(\dfrac{\pi}{8}\right)+\sqrt{a}\right)
\end{split}
\\
\end{align*}
...

1. Integrate[ArcTan[x^4]/(1 + x^4), {x, 0, +Infinity}] // N
2. (-Pi/(2 Sqrt[2])*ArcTan[1]
3.    + Pi/(2 Sqrt[2])*
4.     ArcTan[Power[1, 1/4]*Csc[Pi/8] - Cot[Pi/8]]
5.    + Pi/(2 Sqrt[2])*
6.     ArcTan[Power[1, 1/4]*Csc[Pi/8] + Cot[Pi/8]]
7.    + Pi/(4 Sqrt[2])*
8.     Log[1 - 2*Power[1, 1/4]*Sin[Pi/8] + Sqrt[1]]
9.    - Pi/(4 Sqrt[2])*
10.     Log[1 + 2*Power[1, 1/4]*Sin[Pi/8] + Sqrt[1]]) // N

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...
\begin{align*}
G(1)&=\int_0^{+\infty}\dfrac{\arctan\left(x^4\right)}{1+x^4}\mathrm{d}x\\
G(1)&=\int_0^{+\infty}\dfrac{\arctan\left((bw)^4\right)}{1+(bw)^4}\mathrm{d}(bw)\\
\dfrac{G(1)}{b}&=\int_0^{+\infty}\dfrac{\arctan\left(b^4w^4\right)}{1+b^4w^4}\mathrm{d}w\\
\\
&=\quad
\begin{split}
-\dfrac{\pi}{2\sqrt{2\sqrt{b}\,}}\arctan\left(1\right)\qquad\qquad\qquad\quad\,\,\,\,\,\\
+\dfrac{\pi}{2\sqrt{2\sqrt{b}\,}}\arctan\left(\csc\left(\dfrac{\pi}{8}\right)-\cot\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{2\sqrt{2\sqrt{b}\,}}\arctan\left(\csc\left(\dfrac{\pi}{8}\right)+\cot\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{4\sqrt{2\sqrt{b}\,}}\ln\left(1-\sin\left(\dfrac{\pi}{8}\right)\right)\qquad\quad\\
-\dfrac{\pi}{4\sqrt{2\sqrt{b}\,}}\ln\left(1+\sin\left(\dfrac{\pi}{8}\right)\right)\qquad\quad
\end{split}\\
\\
&=\dfrac{\pi\left(\pi-4\operatorname{artanh}\left(\sin\dfrac{\pi}{8}\right)\right)}{8\sqrt{2\sqrt{b}\,}}\\
&=\dfrac{\pi\left(\pi-4\operatorname{arcoth}\left(\csc\dfrac{\pi}{8}\right)\right)}{8\sqrt{2\sqrt{b}\,}}
\end{align*}
...

1. NIntegrate[ArcTan[2 x^4]/(1 + 2 x^4), {x, 0, +Infinity}] // N
2. (-Pi/(2 Sqrt[2 Sqrt[2]])*ArcTan[1]
3.    + Pi/(2 Sqrt[2 Sqrt[2]])*ArcTan[Csc[Pi/8] - Cot[Pi/8]]
4.    + Pi/(2 Sqrt[2 Sqrt[2]])*ArcTan[Csc[Pi/8] + Cot[Pi/8]]
5.    + Pi/(4 Sqrt[2 Sqrt[2]])*Log[1 - Sin[Pi/8]]
6.    - Pi/(4 Sqrt[2 Sqrt[2]])*Log[1 + Sin[Pi/8]]) // N
7. Pi (Pi - 4 ArcCoth[Csc[Pi/8]])/(8 Sqrt[2 Sqrt[2]]) // N

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青青子衿

回复 2# dim
\begin{align*}
\int_0^{+\infty}\dfrac{\arctan\left(a\,x^2\right)}{1+x^2}\mathrm{d}x&=\qquad
\begin{split}
-\,\dfrac{\pi}{2}\arctan\left(a\right)\qquad\quad\,\,\\
+\,\dfrac{\pi}{2}\arctan\left(\sqrt{2a}+1\right)\\
+\,\dfrac{\pi}{2}\arctan\left(\sqrt{2a}-1\right)\\
\end{split}\\
\\
\int_0^{+\infty}\dfrac{\arctan\left(a\,x^4\right)}{1+x^2}\mathrm{d}x&=\qquad
\begin{split}
\dfrac{\pi}{2}\arctan\left(a\right)\qquad\qquad\qquad\qquad\quad\,\,\,\,\\
-\dfrac{\pi}{2}\arctan\left(\sqrt[4]{a}\csc\left(\dfrac{\pi}{8}\right)-\cot\left(\dfrac{\pi}{8}\right)\right)\\
-\dfrac{\pi}{2}\arctan\left(\sqrt[4]{a}\csc\left(\dfrac{\pi}{8}\right)+\cot\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{2}\arctan\left(\sqrt[4]{a}\sec\left(\dfrac{\pi}{8}\right)-\tan\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{2}\arctan\left(\sqrt[4]{a}\sec\left(\dfrac{\pi}{8}\right)+\tan\left(\dfrac{\pi}{8}\right)\right)
\end{split}\\
\\
\int_0^{+\infty}\dfrac{\arctan\left(a\,x^2\right)}{1+x^4}\mathrm{d}x&=\qquad
\begin{split}
\dfrac{\pi}{2\sqrt2}\arctan\left(\sqrt{a}\,\right)\quad\,\,\,\,\\
+\dfrac{\pi}{4\sqrt2}\ln\left(a+1\right)\quad\\
-\dfrac{\pi}{2\sqrt2}\ln\left(\sqrt{a}+1\right)\\
\end{split}\\
\\
\int_0^{+\infty}\dfrac{\arctan\left(a\,x^4\right)}{1+x^4}\mathrm{d}x&=\qquad
\begin{split}
-\dfrac{\pi}{2\sqrt2}\arctan\left(a\right)\qquad\qquad\qquad\qquad\quad\,\,\,\,\\
+\dfrac{\pi}{2\sqrt2}\arctan\left(\sqrt[4]{a}\csc\left(\dfrac{\pi}{8}\right)-\cot\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{2\sqrt2}\arctan\left(\sqrt[4]{a}\csc\left(\dfrac{\pi}{8}\right)+\cot\left(\dfrac{\pi}{8}\right)\right)\\
+\dfrac{\pi}{4\sqrt2}\ln\left(\sqrt{a}-2\sqrt[4]{a}\sin\left(\dfrac{\pi}{8}\right)+1\right)\\
-\dfrac{\pi}{4\sqrt2}\ln\left(\sqrt{a}+2\sqrt[4]{a}\sin\left(\dfrac{\pi}{8}\right)+1\right)
\end{split}\\
\end{align*}