Question Number 221415 by wewji12 last updated on 04/Jun/25
![∫ dz [−(1/π)((2/z))^ν ∙Σ_(k=0) ^(ν−1) ((Γ(ν−k))/(k!))((z/2))^(2k) +(2/π)ln((1/2)z)J_ν (z)−(1/π)((z/2))^ν Σ_(k=0) ^∞ (((−1)^k (ψ^((0)) (k+ν+1)+ψ^((0)) (k+1)))/(k!(k+ν)!))((z/2))^(2k) ]](https://www.tinkutara.com/question/Q221415.png)
$$\int\:\:\mathrm{d}{z}\:\left[−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\:\frac{\Gamma\left(\nu−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\frac{\mathrm{2}}{\pi}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right){J}_{\nu} \left({z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} \left(\psi^{\left(\mathrm{0}\right)} \left({k}+\nu+\mathrm{1}\right)+\psi^{\left(\mathrm{0}\right)} \left({k}+\mathrm{1}\right)\right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \right] \\ $$
Answered by MrGaster last updated on 05/Jun/25
![=(1/π)∫dz[−((2/z))^ν Σ_(k=0) ^(ν−1) ((Γ(ν−k))/(k!))((z/2))^(2k) +2 ln((z/2))J_ν (z)−((z/2))^ν Σ_(k=0) ^∞ (((−1)^k ψ^((0)) (k+ν+1))/(k!(k+ν)))((z/2))^(2k) −2((z/2))^ν Σ_(k=0) ^∞ (((−1)^k ψ^((0)) (k+1))/(k!(k+ν)!))((z/2))^(2k) ] J_ν (z)=((z/2))^ν Σ_(k=0) ^∞ (((−1)^k )/(k!Γ(k+ν+1)))((z/2))^(2k) ψ^((0)) (n)=−γ+H_(n−1) ,ψ^((0)) (k+ν+1)=−γ+H_(k+ν) ψ^((0)) (k+1)=−γ+H_k ∫dz[−((2/z))^ν Σ_(k=0) ^(ν−1) ((Γ(u−k))/(k!))((z/2))^(2k) +2 ln 2((z/2))((z/2))^ν Σ_(k=0) ^∞ (((−1)^k )/(k!Γ(k+ν+1)))((z/2))^(2k) −((z/2))^ν Σ_(k=0) ^∞ (((−1)^k (−γ+H_(k+ν) ))/(k!(k+ν)!))((z/2))^(2k) −2((z/2))^ν Σ_(k=0) ^∞ (((−1)^k (−γ+H_k ))/(k!(k+ν)!))((z/2))^(2k) ] γΣ_(k=0) ^∞ (((−1)^k )/(k!(k+ν)!))((z/2))^(2k) +γΣ_(k=0) ^∞ (((−1)^k )/(k!(k+ν)))((z/2))^(2k) −H_k Σ_(k=0) ^∞ (((−1)^k )/(k!(k+ν)!))((z/2))^(2k) Σ_(k=0) ^∞ (((−1)^k )/(k!(k+ν)!))((z/2))^(2k) [2γ−H_k −H_(k+ν) +2 ln((z/2))]−((2/z))^ν Σ_(k=0) ^(ν−1) ((Γ(ν−k))/(k!))((z/2))^(2k) H_k +H_(k+ν) =H_k +Σ_(m=1) ^ν (1/(k+m)):H_k +H_(k+ν) =H_k +(H_k +Σ_(m=1) ^ν (1/(k+m)))=2H_k +Σ_(m=1) ^ν (1/(k+m)) Σ_(k=0) ^∞ (((−1)^k )/(k!(k+ν)))((z/2))^(2k) [2γ−2H_k −Σ_(m=1) ^ν (1/(k+m))+2 ln((z/2))] Σ_(k=0) ^(ν−1) (((−1)^k )/(k!(k+ν)!))((z/2))^(2k) [2γ−2H_k −Σ_(m=1) ^ν (1/(k+m))+2 ln((z/2))]+Σ_(k=ν) ^∞ …] lim_(R→∞) ∮_(∣z∣=R) [Σ_(k=ν) ^∞ (((−1)^k )/(k!(k+ν)!))((z/2))^(2k) (2γ−2H_k −Σ_(m=1) ^ν (1/(k+m))+2 ln((z/2)))]dz=0](https://www.tinkutara.com/question/Q221437.png)
$$=\frac{\mathrm{1}}{\pi}\int{dz}\left[−\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\frac{\Gamma\left(\nu−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\mathrm{2}\:\mathrm{ln}\left(\frac{{z}}{\mathrm{2}}\right){J}_{\nu} \left({z}\right)−\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \psi^{\left(\mathrm{0}\right)} \left({k}+\nu+\mathrm{1}\right)}{{k}!\left({k}+\nu\right)}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} −\mathrm{2}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \psi^{\left(\mathrm{0}\right)} \left({k}+\mathrm{1}\right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \right] \\ $$$${J}_{\nu} \left({z}\right)=\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\Gamma\left({k}+\nu+\mathrm{1}\right)}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \\ $$$$\psi^{\left(\mathrm{0}\right)} \left({n}\right)=−\gamma+{H}_{{n}−\mathrm{1}} ,\psi^{\left(\mathrm{0}\right)} \left({k}+\nu+\mathrm{1}\right)=−\gamma+{H}_{{k}+\nu} \\ $$$$\psi^{\left(\mathrm{0}\right)} \left({k}+\mathrm{1}\right)=−\gamma+{H}_{{k}} \\ $$$$\int{dz}\left[−\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\frac{\Gamma\left({u}−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\left(\frac{{z}}{\mathrm{2}}\right)\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\Gamma\left({k}+\nu+\mathrm{1}\right)}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} −\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \left(−\gamma+{H}_{{k}+\nu} \right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} −\mathrm{2}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \left(−\gamma+{H}_{{k}} \right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \right] \\ $$$$\gamma\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\gamma\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({k}+\nu\right)}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} −{H}_{{k}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \left[\mathrm{2}\gamma−{H}_{{k}} −{H}_{{k}+\nu} +\mathrm{2}\:\mathrm{ln}\left(\frac{{z}}{\mathrm{2}}\right)\right]−\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\frac{\Gamma\left(\nu−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \\ $$$${H}_{{k}} +{H}_{{k}+\nu} ={H}_{{k}} +\underset{{m}=\mathrm{1}} {\overset{\nu} {\sum}}\frac{\mathrm{1}}{{k}+{m}}:{H}_{{k}} +{H}_{{k}+\nu} ={H}_{{k}} +\left({H}_{{k}} +\underset{{m}=\mathrm{1}} {\overset{\nu} {\sum}}\frac{\mathrm{1}}{{k}+{m}}\right)=\mathrm{2}{H}_{{k}} +\underset{{m}=\mathrm{1}} {\overset{\nu} {\sum}}\frac{\mathrm{1}}{{k}+{m}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({k}+\nu\right)}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \left[\mathrm{2}\gamma−\mathrm{2}{H}_{{k}} −\underset{{m}=\mathrm{1}} {\overset{\nu} {\sum}}\frac{\mathrm{1}}{{k}+{m}}+\mathrm{2}\:\mathrm{ln}\left(\frac{{z}}{\mathrm{2}}\right)\right] \\ $$$$\left.\underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \left[\mathrm{2}\gamma−\mathrm{2}{H}_{{k}} −\underset{{m}=\mathrm{1}} {\overset{\nu} {\sum}}\frac{\mathrm{1}}{{k}+{m}}+\mathrm{2}\:\mathrm{ln}\left(\frac{{z}}{\mathrm{2}}\right)\right]+\underset{{k}=\nu} {\overset{\infty} {\sum}}\ldots\right] \\ $$$$\underset{{R}\rightarrow\infty} {\mathrm{lim}}\oint_{\mid{z}\mid={R}} \left[\underset{{k}=\nu} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} }{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \left(\mathrm{2}\gamma−\mathrm{2}{H}_{{k}} −\underset{{m}=\mathrm{1}} {\overset{\nu} {\sum}}\frac{\mathrm{1}}{{k}+{m}}+\mathrm{2}\:\mathrm{ln}\left(\frac{{z}}{\mathrm{2}}\right)\right)\right]{dz}=\mathrm{0} \\ $$