パワーな定積分

はじめに

今回はとある依頼を受けて次の定積分を計算したので，その結果を過程と共にここに記そうと思います(といっても，ただひたすら原始関数を求めて逐次積分するだけなので，斬新な発想とかはほとんどありません…)．

因みにこの定積分は，「1辺の長さが1の正方形平板2枚を間隔1で平行に置いたときの形態係数」という値を導出する際に表れます．本題からそれるので，ここで形態係数については説明しません．詳しくは Wikipedia 等を参照して頂きたいと思います．

解答

早速計算を始めます．被積分関数の分母はどの変数についても複2次多項式なので，どれでも積分できますが，ここではとりあえず$y_2$で積分します．

$y_2$で積分

コチラの記事 内にある複2次多項式の逆数の積分より，
$$\begin{array}{rl}& {\int }_0^1\frac{d{y}_2}{\{(x_2-x_1{)}^2+(y_2-y_1{)}^2+1{\}}^2}\\ & =\frac{1}{2\{(x_2-x_1{)}^2+1{\}}^{\frac{3}{2}}}{[\frac{\sqrt{(x_2-x_1{)}^2+1}(y_2-y_1)}{(x_2-x_1{)}^2+1+(y_2-y_1{)}^2}+\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{y_2-y_1}{\sqrt{(x_2-x_1{)}^2+1}}\right)]}_0^1\\ & =\frac{1}{2\{(x_2-x_1{)}^2+1{\}}^{\frac{3}{2}}}\{\frac{\sqrt{(x_2-x_1{)}^2+1}(1-y_1)}{(x_2-x_1{)}^2+1+(1-y_1{)}^2}+\frac{\sqrt{(x_2-x_1{)}^2+1}{y}_1}{(x_2-x_1{)}^2+1+y_1^2}\}\\ & +\frac{1}{2\{(x_2-x_1{)}^2+1{\}}^{\frac{3}{2}}}\{\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1-y_1}{\sqrt{(x_2-x_1{)}^2+1}}\right)+\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{y_1}{\sqrt{(x_2-x_1{)}^2+1}}\right)\}\end{array}$$
となります．

$y_1$で積分

一気に被積分関数がぐちゃぐちゃになりましたが，じっと眺めていると$y_1$に関しては比較的簡単な形をしていることがおぼろげながら見えてくると思います．そこで次は$y_1$で積分することにします．すると最初の2項が
$$\begin{array}{rl}{\int }_0^1\frac{\sqrt{(x_2-x_1{)}^2+1}{y}_1}{(x_2-x_1{)}^2+1+y_1^2}d{y}_1& =\frac{\sqrt{(x_2-x_1{)}^2+1}}{2}{[\log |(x_2-x_1{)}^2+1+y_1^2|]}_0^1\\ & =\frac{\sqrt{(x_2-x_1{)}^2+1}}{2}\log \left(\frac{(x_2-x_1{)}^2+2}{(x_2-x_1{)}^2+1}\right)\end{array}$$
$$\begin{array}{rl}{\int }_0^1\frac{\sqrt{(x_2-x_1{)}^2+1}(1-y_1)}{(x_2-x_1{)}^2+1+(1-y_1{)}^2}d{y}_1& ={\int }_1^0\frac{\sqrt{(x_2-x_1{)}^2+1}z}{(x_2-x_1{)}^2+1+z^2}(-dz)(z=1-y_1とおいた)\\ & ={\int }_0^1\frac{\sqrt{(x_2-x_1{)}^2+1}z}{(x_2-x_1{)}^2+1+z^2}dz\\ & =\frac{\sqrt{(x_2-x_1{)}^2+1}}{2}\log \left(\frac{(x_2-x_1{)}^2+2}{(x_2-x_1{)}^2+1}\right)\end{array}$$
と計算されます．後ろの2項は コチラの記事 内の$\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(ax+b)$の積分より，
$$\begin{array}{rl}{\int }_0^1\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{y_1}{\sqrt{(x_2-x_1{)}^2+1}}\right)d{y}_1& ={[y_1\cdot \mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{y_1}{\sqrt{(x_2-x_1{)}^2+1}}\right)-\frac{\sqrt{(x_2-x_1{)}^2+1}}{2}\log |1+{\left(\frac{y_1}{\sqrt{(x_2-x_1{)}^2+1}}\right)}^2|]}_0^1\\ & =\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{(x_2-x_1{)}^2+1}}\right)-\frac{\sqrt{(x_2-x_1{)}^2+1}}{2}\log \left(\frac{(x_2-x_1{)}^2+2}{(x_2-x_1{)}^2+1}\right)\end{array}$$
$$\begin{array}{rl}{\int }_0^1\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1-y_1}{\sqrt{(x_2-x_1{)}^2+1}}\right)d{y}_1& ={\int }_1^0\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{z}{\sqrt{(x_2-x_1{)}^2+1}}\right)(-dz)(z=1-y_1とおいた)\\ & ={\int }_0^1\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{z}{\sqrt{(x_2-x_1{)}^2+1}}\right)dz\\ & =\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{(x_2-x_1{)}^2+1}}\right)-\frac{\sqrt{(x_2-x_1{)}^2+1}}{2}\log \left(\frac{(x_2-x_1{)}^2+2}{(x_2-x_1{)}^2+1}\right)\end{array}$$

と計算できます．従って，
$$\begin{array}{r}{\int }_0^{1}d{y}_1{\int }_0^{1}d{y}_2\frac{1}{\{(x_2-x_1{)}^2+(y_2-y_1{)}^2+1{\}}^2}=\frac{1}{\{(x_2-x_1{)}^2+1{\}}^{\frac{3}{2}}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{(x_2-x_1{)}^2+1}}\right)\end{array}$$
が成り立ちます．被積分関数がちょっとスッキリして安心するポイントです．

$x_2$で積分

次は$x_1,x_2$のいずれで積分しても良いですが，とりあえず$x_2$で積分してみます． コチラの記事 内の，分母が2次式の3/2乗の積分を用いて部分積分することで，
$$\begin{array}{rl}& {\int }_0^1\frac{1}{\{(x_2-x_1{)}^2+1{\}}^{\frac{3}{2}}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{(x_2-x_1{)}^2+1}}\right)d{x}_2\\ & ={\left[\frac{x_2-x_1}{\sqrt{1+(x_2-x_1{)}^2}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{(x_2-x_1{)}^2+1}}\right)\right]}_0^1-{\int }_0^1\frac{x_2-x_1}{\sqrt{1+(x_2-x_1{)}^2}}\frac{-(x_2-x_1)}{\{(x_2-x_1{)}^2+1{\}}^{\frac{3}{2}}}\frac{1}{1+{\left(\frac{1}{\sqrt{(x_2-x_1{)}^2+1}}\right)}^2}d{x}_2\\ & =\frac{1-x_1}{\sqrt{1+(1-x_1{)}^2}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{1+(1-x_1{)}^2}}\right)+\frac{x_1}{\sqrt{1+x_1^2}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{1+x_1^2}}\right)\\ & +{\int }_0^1\frac{x_2-x_1}{\sqrt{1+(x_2-x_1{)}^2}}\frac{x_2-x_1}{\{(x_2-x_1{)}^2+1{\}}^{\frac{3}{2}}}\frac{(x_2-x_1{)}^2+1}{(x_2-x_1{)}^2+2}d{x}_2\\ & =\frac{1-x_1}{\sqrt{(1-x_1{)}^2+1}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{(1-x_1{)}^2+1}}\right)+\frac{x_1}{\sqrt{x_1^2+1}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{x_1^2+1}}\right)\\ & +{\int }_0^1\frac{(x_2-x_1{)}^2}{((x_2-x_1{)}^2+1)((x_2-x_1{)}^2+2)}d{x}_2\end{array}$$
が成り立ちます．最後の積分は，
$$\begin{array}{rl}& {\int }_0^1\frac{(x_2-x_1{)}^2}{((x_2-x_1{)}^2+1)((x_2-x_1{)}^2+2)}d{x}_2\\ & ={\int }_0^1\{\frac{-1}{(x_2-x_1{)}^2+1}+\frac{2}{(x_2-x_1{)}^2+2}\}d{x}_2\\ & ={[-\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(x_2-x_1)+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{x_2-x_1}{\sqrt{2}}\right)]}_0^1\\ & =-\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(1-x_1)+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1-x_1}{\sqrt{2}}\right)+\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(-x_1)-\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{-x_1}{\sqrt{2}}\right)\\ & =\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(x_1-1)-\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{x_1-1}{\sqrt{2}}\right)-\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(x_1)+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{x_1}{\sqrt{2}}\right)\end{array}$$
と計算されるので，与式を$y_2,y_1,x_2$で積分した結果は，
$$\begin{array}{rl}& {\int }_0^{1}d{y}_1{\int }_0^{1}d{x}_2{\int }_0^{1}d{y}_2\frac{1}{\{(x_2-x_1{)}^2+(y_2-y_1{)}^2+1{\}}^2}\\ & =\frac{1-x_1}{\sqrt{(1-x_1{)}^2+1}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{(1-x_1{)}^2+1}}\right)+\frac{x_1}{\sqrt{x_1^2+1}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{x_1^2+1}}\right)\\ & -\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(x_1)+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{x_1}{\sqrt{2}}\right)+\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(x_1-1)-\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{x_1-1}{\sqrt{2}}\right)\end{array}$$
となります．

$x_1$で積分

最後に$x_1$で各項を積分しましょう．$(関数)\times \mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}$という形が入っている2つ目の項は先ほどと同様に部分積分を使って，
$$\begin{array}{rl}{\int }_0^1\frac{x_1}{\sqrt{x_1^2+1}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{x_1^2+1}}\right)d{x}_1& ={\left[\sqrt{x_1^2+1}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{x_1^2+1}}\right)\right]}_0^1-{\int }_0^1\sqrt{x_1^2+1}\frac{-x_1}{(x_1^2+1{)}^{\frac{3}{2}}}\frac{x_1^2+1}{x_1^2+2}d{x}_1\\ & =-\frac{\pi }{4}+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)+{\int }_0^1\frac{x_1}{x_1^2+2}d{x}_1\\ & =-\frac{\pi }{4}+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)+{[\frac{1}{2}\log |x_1^2+2|]}_0^1\\ & =-\frac{\pi }{4}+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{2}\log \frac{2}{3}\end{array}$$
と計算できます．これより，
$$\begin{array}{rl}{\int }_0^1\frac{1-x_1}{\sqrt{(1-x_1{)}^2+1}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{(1-x_1{)}^2+1}}\right)d{x}_1& ={\int }_1^0\frac{z}{\sqrt{z^2+1}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{z^2+1}}\right)(-dz)(z=1-x_1とおいた)\\ & ={\int }_0^1\frac{z}{\sqrt{z^2+1}}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{z^2+1}}\right)dz\\ & =-\frac{\pi }{4}+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{2}\log \frac{2}{3}\end{array}$$
も成り立ちます．残りの項については，
$${\int }_0^1\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(x_1)d{x}_1={[x_1\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(x_1)-\frac{1}{2}\log (1+x_1^2)]}_0^1=\frac{\pi }{4}-\frac{1}{2}\log 2$$
$${\int }_0^1\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{x_1}{\sqrt{2}}\right)d{x}_1=\sqrt{2}{[x_1\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{x_1}{\sqrt{2}}\right)-\frac{\sqrt{2}}{2}\log (1+\frac{x_1^2}{2})]}_0^1=\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)-\log \frac{3}{2}$$
$${\int }_0^1\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(x_1-1)d{x}_1\stackrel{z=1-x_1と置換}{=}-{\int }_0^1\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}(z)dz=-\frac{\pi }{4}+\frac{1}{2}\log 2$$
$$\begin{array}{rl}& {\int }_0^1\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{x_1-1}{\sqrt{2}}\right)d{x}_1\stackrel{z=1-x_1と置換}{=}-{\int }_0^1\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{z}{\sqrt{2}}\right)dz=-\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)+\log \frac{3}{2}\end{array}$$
が成り立つので，$x_1$で積分した結果は
$$\begin{array}{rl}& {\int }_0^{1}d{x}_1{\int }_0^{1}d{y}_1{\int }_0^{1}d{x}_2{\int }_0^{1}d{y}_2\frac{1}{\{(x_2-x_1{)}^2+(y_2-y_1{)}^2+1{\}}^2}\\ & =-\frac{\pi }{4}+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{2}\log \frac{2}{3}-\frac{\pi }{4}+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)-\frac{1}{2}\log \frac{2}{3}\\ & -\frac{\pi }{4}+\frac{1}{2}\log 2+\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)-\log \frac{3}{2}-\frac{\pi }{4}+\frac{1}{2}\log 2-\{-\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)+\log \frac{3}{2}\}\\ & =4\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)+\log \frac{4}{3}-\pi \end{array}$$
これが求める値です．そこそこキレイな値が得られました！

この結果を定理として挙げておきます．

実数値

WolframAlphaで計算したところ，
$$4\sqrt{2}\mathrm{A}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\left(\frac{1}{\sqrt{2}}\right)+\log \frac{4}{3}-\pi =0.6277684243\cdots$$
でした.

発展

上と全く同様の轍を踏むことで，次の定積分も計算できます．

実はこの値は，「1辺の長さが$L$の正方形平板2枚を間隔$H$で平行に置いたときの形態係数」を求める際に表れます．$L=H=1$即ち$w=1$を代入すると，上で得た値と等しくなることも確認できます．

さいごに

初めてこの定積分を見たときは，閉じた値が求まりそうになかったので絶望しかけました．が，最終的に閉じた値が得られて良かったです．

今回の記事は以上です．
お読み頂きありがとうございました．