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{\large\textbf{��11�񕡑f��͊wII���K (2011�N6��28�����{)}}
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�S������ ���n����/ TA �v�{�q�V�@���{���F
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�ȉ��̖������̉��K�̎��Ԃ́w�͂��߁x�Ƀ��|�[�g�Ƃ��Ē�o���邱��.


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\noindent
\textbf{[R14]} (1)
$f(z)=\int_0^1 e^{-t}t^{z-1}dt$��$\C$��̗L���^�֐��ɉ�͐ڑ��\�ł���
$$
f(z)=\sum_{n=1}^\infty \frac{(-1)^n}{n!}\frac{1}{z+n}
$$
�����藧���Ƃ������D�i�q���g�F$e^{-t}$�̃e�[���[�W�J���g���j

(2) �K���}�֐�$\Gamma(z)$ (���[45]�Q��)��$f(z)$�̍���$\C$��Ő����ł��邱�Ƃ������D

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�ȉ��͋������\�p�̖��ł��D






\prob  $f(z)$��$\C$���$C^1$�֐���$\Delta$�̊O�ł͍P���I��$0$�ɂȂ���̂Ƃ���D
���̂Ƃ�
$$
u(\zeta)=\iint_\C\frac{f(z)}{z-\zeta}dxdy
$$
��$C^1$�ł���C$\C\setminus\Delta$��ł͐����ł��邱�Ƃ������D

\prob
($\C$�ł�Mittag-Leffler�̒藝)  
���f����$\{z_n\}_{n\ge1 }$��$0<|z_1|\le |z_2|\le |z_3|\le\cdots$,
$\lim_{n\to\infty}|z_n|=\infty$�𖞂����Ƃ���D������$G_n(z)$�ɑ΂��đ�����
$P_n(z)$��
$$
\left| G_n\left(\frac{1}{z-z_n}\right)-P_n(z)\right|<2^{-n} \quad\text{for}\quad
z\in\Delta(0,|z_n|/2)
$$
�𖞂����悤�ɑI�ׂ�
$$
F(z)=\sum_{n\ge 1} \left(G_n\left(\frac{1}{z-z_n}\right)-P_n(z)\right)
$$
�͗L���^�֐��Ɏ������邱�Ƃ������D�܂�$z_1=0$�̏ꍇ�ɂ͂��̍\�����@���ǂ̂悤�ɏC������΂悢���H

\prob
���̂悤�Ȑ���������$\mathbb{C}$��̗L���^�֐�$F(z)$���\������F

(a) $z_n=n$, $n\in\mathbb{N}$�ŗ���$n$�̂P�ʂ̋ɂ����D


(b) $z_n=\sqrt{n}$, $n\in\mathbb{N}$�ŗ���$1$�̂P�ʂ̋ɂ����D



(c) $n\in\mathbb{Z}$�ŗ���$1$�̂P�ʂ̋ɂ����D

(d) $(\mathbb{Z}+i\mathbb{Z})\setminus\{0\}$�ŗ���$1$�̂P�ʂ̋ɂ����D


(e) $\mathbb{Z}+i\mathbb{Z}$�Ŏ�v��$(z-(m+ni))^{-3}$�̋ɂ����D


\prob
����$g\in C^\infty(\C)$�ɑ΂��Ĕ���������$\partial f/\partial \overline{z}=g$�̉�������߂�D

(a) $g(z)=z$, \  

(b) $g(z)=\overline{z}$, \ 

(c) $g(z)=z^n \overline{z}^m, n,m\in \mathbb{N}$, \

(d) $g(z)=\exp \overline{z}$.



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(b) $f(z)$ �͔��������� $f''(z)=6 \{ f(z) \}^2 -4 f(z)$ ��
���������Ƃ������D\\
�i�q���g�F$\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^2} = 
\frac{\pi^2}{6}$, 
$\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^4} = \frac{\pi^4}{90}$ 
�Ȃǂ�p���āC$f(z)$ �� $z=0$ �̂܂��Ń��[�����W�J����D�j

(c) ���̓������ؖ�����D�������C�ϕ��H�͓K���ɑI�Ԃ��̂Ƃ���D
\[ z = \frac{1}{2} \int_{f(z)}^{\infty} \frac{dw}{\sqrt{w^2(w-1)}}. \]