含有对数与根式的二重瑕积分

青青子衿

\[ \color{black}{\int_0^1\int_0^1 \dfrac{\ln\left(x+y\right)}{\sqrt{1+x+y\,}}{\rm\,d}x\!{\rm\,d}y} \]
\begin{array}{c}
\color{black}
-\dfrac{20}{9}+\dfrac{104}{9}\sqrt{2\,}-\dfrac{28}{3}\sqrt{3}-\dfrac{8}{3}\ln2+4\sqrt{3}\ln2\\
-\dfrac{40}{3}\ln\Big(1+\sqrt{2\,}\,\Big)+\dfrac{32}{3}\ln\Big(1+\sqrt{3\,}\,\Big)
\end{array}

hbghlyj

Substitute $\cases{u=x+y\\v=x-y}$. Then, $\cases{x=\frac{u+v}{2}\\y=\frac{u-v}2}$, and the domain of integration become $$(0<u\leq 1\land -u\leq v\leq u)\lor (1<u<2\land u-2\leq v\leq 2-u)$$
From Jacobian, $dxdy = \frac{1}{2}dudv$. The integral becomes $I_1+I_2$
$$I_1=\int_0^1\int_{-u}^u \frac{\ln(u)}{\sqrt{1+u}}\cdot \frac{1}{2}dvdu$$
and
$$I_2=\int_1^2\int_{u-2}^{2-u} \frac{\ln(u)}{\sqrt{1+u}}\cdot \frac{1}{2}dvdu$$
Integrate wrt $v$:
$$I_1=\int_0^1 \frac{u\ln(u)}{\sqrt{1+u}}du$$


and
$$I_2=\int_1^2 \frac{(2-u)\ln(u)}{\sqrt{1+u}}du$$

hbghlyj

\begin{align*}
& \int_{1}^{2} \frac{(2-u)\log(u)}{\sqrt{u+1}}\,du\\
&\text{Substitute } s=\sqrt{u+1} \text{ and } ds=\frac{1}{2\sqrt{u+1}}du. \\
&\text{This gives a new lower bound } s=\sqrt{1+1}=\sqrt{2} \text{ and upper bound } s=\sqrt{1+2}=\sqrt{3}:\\
&= -2 \int_{\sqrt{2}}^{\sqrt{3}} (s^2-3)\log(s^2-1)\,ds \\
&\text{Expanding the integrand } (s^2-3)\log(s^2-1) \text{ gives } s^2\log(s^2-1)-3\log(s^2-1):\\
&= -2 \int_{\sqrt{2}}^{\sqrt{3}} (s^2\log(s^2-1)-3\log(s^2-1))\,ds \\
&\text{Integrate the sum term by term and factor out constants: }\\
&= -2 \int_{\sqrt{2}}^{\sqrt{3}} s^2\log(s^2-1)\,ds + 6 \int_{\sqrt{2}}^{\sqrt{3}} \log(s^2-1)\,ds \\
&\text{For the integrand } s^2\log(s^2-1), \text{ integrate by parts, } \int f dg = fg - \int g df,\\
&\text{where } f=\log(s^2-1), dg=s^2ds, df=\frac{2s}{s^2-1}ds, g=\frac{s^3}{3}:\\
&= \left(-\frac{2}{3}s^3\log(s^2-1)\right)\Biggr|_{\sqrt{2}}^{\sqrt{3}} + \frac{2}{3}\int_{\sqrt{2}}^{\sqrt{3}}\frac{2s^4}{s^2-1}\,ds + 6 \int_{\sqrt{2}}^{\sqrt{3}} \log(s^2-1)\,ds \\&= \left(-\frac{2}{3}\sqrt{3}^3\log(\sqrt{3}^2-1)\right)-\left(-\frac{2}{3}\sqrt{2}^3\log(\sqrt{2}^2-1)\right)+ \frac{4}{3}\int_{\sqrt{2}}^{\sqrt{3}}\frac{s^4}{s^2-1}\,ds + 6 \int_{\sqrt{2}}^{\sqrt{3}} \log(s^2-1)\,ds \\
&= -\sqrt{3}\log(4) + \frac{4}{3}\int_{\sqrt{2}}^{\sqrt{3}}\frac{s^4}{s^2-1}\,ds + 6 \int_{\sqrt{2}}^{\sqrt{3}} \log(s^2-1)\,ds \\
&\text{For the integrand }\frac{s^4}{s^2-1}\text{, do long division: }\\
&= -\sqrt{3} \log(4) + \frac{4}{3} \int_{\sqrt{2}}^{\sqrt{3}} \left(s^2 + \frac{1}{2(s-1)} - \frac{1}{2(s+1)} + 1\right) ds + 6 \int_{\sqrt{2}}^{\sqrt{3}} \log(s^2-1) ds\\
&\text{Apply the fundamental theorem of calculus.}\\
&= -\frac{8\sqrt{2}}{9} + \frac{4}{\sqrt{3}} + \frac{4}{3}(\sqrt{3}-\sqrt{2}) - \sqrt{3}\log(4) - \frac{2}{3}(\log(\sqrt{2}-1)-\log(\sqrt{3}-1)) \\
&\quad + \frac{2}{3}(\log(1+\sqrt{2})-\log(1+\sqrt{3})) + 6\int_{\sqrt2}^{\sqrt{3}}\log(s^2-1)\,ds \\
&\text{For the integrand } \log(s^2-1), \text{integrate by parts,} \\
&\int f\,dg = fg - \int g\,df \text{, where } f = \log(s^2-1), dg = ds, df = {2 s\over s^2 - 1} ds, g = s:\\
&= -\frac{8\sqrt{2}}{9} + \frac{4}{\sqrt{3}} + \frac{4}{3}(\sqrt{3}-\sqrt{2}) - \sqrt{3}\log(4) - \frac{2}{3}(\log(\sqrt{2}-1)-\log(\sqrt{3}-1))\\
&\quad+ \frac{2}{3}(\log(1+\sqrt{2})-\log(1+\sqrt{3})) + 6s\log(s^2-1)\Big|_{\sqrt{2}}^{\sqrt{3}} - 6\int_{\sqrt{2}}^{\sqrt{3}}\frac{2s^2}{s^2-1}ds\\
&\text{Evaluate the antiderivative at the limits and subtract.}\\&\qquad6s\log(s^2-1)\Big|_{\sqrt{2}}^{\sqrt{3}}= 6\sqrt{3}\log(\sqrt{3}^2-1) - 6\sqrt{2}\log(\sqrt{2}^2-1)= \sqrt{3}\log(64)\\
&= -\frac{8\sqrt{2}}{9} + \frac{4}{\sqrt{3}} + \frac{4}{3}(\sqrt{3}-\sqrt{2}) - \sqrt{3}\log(4) + \sqrt{3}\log(64) - \frac{2}{3}(\log(\sqrt{2}-1)-\log(\sqrt{3}-1))\\
&\quad+ \frac{2}{3}(\log(1+\sqrt{2})-\log(1+\sqrt{3})) - 12\int_{\sqrt{2}}^{\sqrt{3}}\frac{s^2}{s^2-1}ds
\\&\text{For the integrand }\frac{s^2}{s^2 - 1}\text{, do long division}:\\
&= -\frac{8\sqrt{2}}{9} + \frac{4}{\sqrt{3}} + \frac{4}{3}(\sqrt{3}-\sqrt{2}) - \sqrt{3}\log(4) + \sqrt{3}\log(64) - \frac{2}{3}(\log(\sqrt{2}-1)-\log(\sqrt{3}-1))\\
&\quad+ \frac{2}{3}(\log(1+\sqrt{2})-\log(1+\sqrt{3})) - 12\int_{\sqrt{2}}^{\sqrt{3}}\left(-\frac{1}{2(s+1)} + \frac{1}{2(s-1)} + 1\right)ds\\
&\text{Apply the fundamental theorem of calculus.}\\
&= -\frac{8\sqrt{2}}{9} + \frac{4}{\sqrt{3}} + 12(\sqrt{2}-\sqrt{3}) + \frac{4}{3}(\sqrt{3}-\sqrt{2}) - \sqrt{3}\log(4) + \sqrt{3}\log(64)\\
&\quad - 6\log(1+\sqrt{2}) + \frac{16}{3}(\log(\sqrt{2}-1)-\log(\sqrt{3}-1)) + \frac{2}{3}(\log(1+\sqrt{2})-\log(1+\sqrt{3})) + 6\log(1+\sqrt{3})\\
&= \frac{2}{9}(44\sqrt{2}-42\sqrt{3}+9\sqrt{3}\log(4)-48\tanh^{-1}(\sqrt{2})+48\tanh^{-1}(\sqrt{3}))\\
\end{align*}

hbghlyj

Get LaTeX form from step by step solution
\begin{align*}&\text{Take the integral:} \\&
f(u)=\int \frac{u \log (u)}{\sqrt{u+1}} \, du \\& \text{For }\text{the }\text{integrand }\frac{u \log (u)}{\sqrt{u+1}}, \text{substitute }s=\sqrt{u+1} \text{and }ds=\frac{1}{2 \sqrt{u+1}}\, du: \\&
  =\int 2 \left(s^2-1\right) \log \left(s^2-1\right) \, ds \\& \text{Factor }\text{out }\text{constants:} \\&
  =2\int \left(s^2-1\right) \log \left(s^2-1\right) \, ds \\& \text{Expanding }\text{the }\text{integrand }\left(s^2-1\right) \log \left(s^2-1\right) \text{gives }s^2 \log \left(s^2-1\right)-\log \left(s^2-1\right): \\&
  =2\int \left(s^2 \log \left(s^2-1\right)-\log \left(s^2-1\right)\right) \, ds \\& \text{Integrate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants:} \\&
  =2\int s^2 \log \left(s^2-1\right) \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{For }\text{the }\text{integrand }s^2 \log \left(s^2-1\right), \text{integrate }\text{by }\text{parts, }\int f \, dg=f g-\int g \, df, \text{where }\\&f=\log \left(s^2-1\right),    dg=s^2\, ds,\\&df=\frac{2 s}{s^2-1}\, ds,     g=\frac{s^3}{3}: \\&
  =\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2}{3}\int \frac{2 s^4}{s^2-1} \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{Factor }\text{out }\text{constants:} \\&
  =\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{4}{3}\int \frac{s^4}{s^2-1} \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{For }\text{the }\text{integrand }\frac{s^4}{s^2-1}, \text{do }\text{long }\text{division:} \\&
  =\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{4}{3}\int \left(s^2+\frac{1}{2 (s-1)}-\frac{1}{2 (s+1)}+1\right) \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{Integrate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants:} \\&
  =\frac{2}{3} s^3 \log \left(s^2-1\right)+\frac{2}{3}\int \frac{1}{s+1} \, ds-\frac{4}{3}\int s^2 \, ds-\frac{2}{3}\int \frac{1}{s-1} \, ds-\frac{4}{3}\int 1 \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{For }\text{the }\text{integrand }\frac{1}{s+1}, \text{substitute }p=s+1 \text{and }dp=\, ds: \\&
  =\frac{2}{3} s^3 \log \left(s^2-1\right)+\frac{2}{3}\int \frac{1}{p} \, dp-\frac{4}{3}\int s^2 \, ds-\frac{2}{3}\int \frac{1}{s-1} \, ds-\frac{4}{3}\int 1 \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{The }\text{integral }\text{of }\frac{1}{p} \text{ is }\log (p): \\&
  =\frac{2 \log (p)}{3}+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{4}{3}\int s^2 \, ds-\frac{2}{3}\int \frac{1}{s-1} \, ds-\frac{4}{3}\int 1 \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{The }\text{integral }\text{of }s^2 \text{ is }\frac{s^3}{3}: \\&
  =-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2}{3}\int \frac{1}{s-1} \, ds-\frac{4}{3}\int 1 \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{For }\text{the }\text{integrand }\frac{1}{s-1}, \text{substitute }w=s-1 \text{and }dw=\, ds: \\&
  =-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2}{3}\int \frac{1}{w} \, dw-\frac{4}{3}\int 1 \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{The }\text{integral }\text{of }\frac{1}{w} \text{ is }\log (w): \\&
  =-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2 \log (w)}{3}-\frac{4}{3}\int 1 \, ds-2\int \log \left(s^2-1\right) \, ds \\& \text{The }\text{integral }\text{of }1 \text{ is }s: \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2 \log (w)}{3}-2\int \log \left(s^2-1\right) \, ds \\& \text{For }\text{the }\text{integrand }\log \left(s^2-1\right), \text{integrate }\text{by }\text{parts, }\int f \, dg=f g-\int g \, df, \text{where }\\&f=\log \left(s^2-1\right),    dg=\, ds,\\&df=\frac{2 s}{s^2-1}\, ds,     g=s: \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2 \log (w)}{3}+2\int \frac{2 s^2}{s^2-1} \, ds \\& \text{Factor }\text{out }\text{constants:} \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2 \log (w)}{3}+4\int \frac{s^2}{s^2-1} \, ds \\& \text{For }\text{the }\text{integrand }\frac{s^2}{s^2-1}, \text{do }\text{long }\text{division:} \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2 \log (w)}{3}+4\int \left(-\frac{1}{2 (s+1)}+\frac{1}{2 (s-1)}+1\right) \, ds \\& \text{Integrate }\text{the }\text{sum }\text{term }\text{by }\text{term }\text{and }\text{factor }\text{out }\text{constants:} \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2 \log (w)}{3}-2\int \frac{1}{s+1} \, ds+2\int \frac{1}{s-1} \, ds+4\int 1 \, ds \\& \text{For }\text{the }\text{integrand }\frac{1}{s+1}, \text{substitute }v=s+1 \text{and }dv=\, ds: \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)-\frac{2 \log (w)}{3}-2\int \frac{1}{v} \, dv+2\int \frac{1}{s-1} \, ds+4\int 1 \, ds \\& \text{The }\text{integral }\text{of }\frac{1}{v} \text{ is }\log (v): \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)-2 \log (v)-\frac{2 \log (w)}{3}+2\int \frac{1}{s-1} \, ds+4\int 1 \, ds \\& \text{For }\text{the }\text{integrand }\frac{1}{s-1}, \text{substitute }z_1=s-1 \text{and }dz_1=\, ds: \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)-2 \log (v)-\frac{2 \log (w)}{3}+2\int \frac{1}{z_1} \, dz_1+4\int 1 \, ds \\& \text{The }\text{integral }\text{of }\frac{1}{z_1} \text{ is }\log \left(z_1\right): \\&
  =-\frac{4 s}{3}-\frac{4 s^3}{9}+\frac{2 \log (p)}{3}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)-2 \log (v)-\frac{2 \log (w)}{3}+2 \log \left(z_1\right)+4\int 1 \, ds \\& \text{The }\text{integral }\text{of }1 \text{ is }s: \\&
  =\frac{2 \log (p)}{3}-\frac{4 s^3}{9}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)+\frac{8 s}{3}-2 \log (v)-\frac{2 \log (w)}{3}+2 \log \left(z_1\right)+\text{constant} \\& \text{Substitute }\text{back }\text{for }z_1=s-1: \\&
  =\frac{2 \log (p)}{3}-\frac{4 s^3}{9}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)+\frac{8 s}{3}+2 \log (s-1)-2 \log (v)-\frac{2 \log (w)}{3}+\text{constant} \\& \text{Substitute }\text{back }\text{for }v=s+1: \\&
  =\frac{2 \log (p)}{3}-\frac{4 s^3}{9}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)+\frac{8 s}{3}+2 \log (s-1)-2 \log (s+1)-\frac{2 \log (w)}{3}+\text{constant} \\& \text{Substitute }\text{back }\text{for }w=s-1: \\&
  =\frac{2 \log (p)}{3}-\frac{4 s^3}{9}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)+\frac{8 s}{3}+\frac{4}{3} \log (s-1)-2 \log (s+1)+\text{constant} \\& \text{Substitute }\text{back }\text{for }p=s+1: \\&
  =-\frac{4 s^3}{9}-2 s \log \left(s^2-1\right)+\frac{2}{3} s^3 \log \left(s^2-1\right)+\frac{8 s}{3}+\frac{4}{3} \log (s-1)-\frac{4}{3} \log (s+1)+\text{constant} \\& \text{Substitute }\text{back }\text{for }s=\sqrt{u+1}: \\&
  =-\frac{4}{9} (u+1)^{3/2}+\frac{8 \sqrt{u+1}}{3}+\frac{2}{3} (u+1)^{3/2} \log (u)-2 \sqrt{u+1} \log (u)+\frac{4}{3} \log \left(\sqrt{u+1}-1\right)-\frac{4}{3} \log \left(\sqrt{u+1}+1\right)+\text{constant} \\& \text{Evaluate at }u=0\text{ and }u=1\\&I_1=f(1)-f(0)=\frac{2}{9} \left(8 \sqrt{2}-6 \left(\log \left(\sqrt{2}+1\right)-\log \left(\sqrt{2}-1\right)\right)\right)-\frac{2}{9} (10-12 \log (2))=\frac{4}{9} \left(4 \sqrt{2}-5+\log (64)-6 \sinh ^{-1}(1)\right)
\end{align*}

hbghlyj

$$\int_0^1 \int_0^1\frac{\ln\left(x^{2}+y^{2}\right)}{\sqrt{x+y}}dx dy$$math.stackexchange.com/questions/3403701
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