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Openai/6958fa17-8214-800f-b483-22f11acd2dc2
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==== 4. 複素解析側の中身(要点だけ) ==== MSEにある Hardy(由来はRamanujan)の流れと同様に、Abel–Planaの補正項を # 反射公式(または sin と Γ の関係)で整理して、t を -∞〜∞ に拡張した形の積分に直す # そこで出てくる e^{-\pi|t|} を上の恒等式で u 積分に置き換える # 積分順序を交換し、内側の t 積分を Mellin 変換の逆変換(輪郭積分)として評価する という手順を踏むと、結局 −i∫0∞f(it)−f(−it)e2πt−1 dt=a∫0∞e−auln2u+π2 du-i\int_{0}^{\infty}\frac{f(it)-f(-it)}{e^{2\pi t}-1}\,dt = a\int_{0}^{\infty}\frac{e^{-a u}}{\ln^{2}u+\pi^{2}}\,du−i∫0∞e2πt−1f(it)−f(−it)dt=a∫0∞ln2u+π2e−audu が得られる。ここで最後の “内側の t 積分が e^{-a u} に落ちる” 部分は、MSE解答中で「Mellin transform gives … = -e^{-u}」として書かれている箇所の a 付き一般化である。 Mathematics Stack Exchange<ref>{{cite web|title=Mathematics Stack Exchange|url=https://math.stackexchange.com/questions/741014/prove-int-0-infty-frac1-gammax-mathrmdx-e-int-0-infty-fr|publisher=Mathematics Stack Exchange|access-date=2026-01-04}}</ref> (厳密に書くなら、Γ と sin の関係を使って 1/Γ(z) の Hankel 型表現や、縦線 Re(z)=0 上の積分を閉曲線で評価して留数を拾う、という標準手順になる。Abel–Plana自体が複素解析由来なので、この段階は複素解析が避けにくい。 ウィキペディア<ref>{{cite web|title=ウィキペディア|url=https://en.wikipedia.org/wiki/Abel%E2%80%93Plana_formula|publisher=en.wikipedia.org|access-date=2026-01-04}}</ref>)
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