自作1

$$ \int_{0}^{1}t^{k+m}dt=\frac{1}{k+m+1} $$
であるから、
$$\sum_{k=0}^{n} (-1)^{k}\binom{n}{k} \frac{1}{k+m+1} \\ =\sum_{k=0}^{n} (-1)^{k}\binom{n}{k}\int_{0}^{1}t^{k+m}dt \\ =\int_{0}^{1}\sum_{k=0}^{n} (-1)^{k}\binom{n}{k}t^{k+m}dt \\ =\int_{0}^{1}\sum_{k=0}^{n} (-t)^{k}\binom{n}{k}t^{m}dt \\ =\int_{0}^{1}(1-t)^nt^{m}dt\\ $$
ここで、$1-t=s$と置換して
$$ =\int_{0}^{1}s^n(1-s)^{m}ds\\ =\int_{0}^{1}s^n\sum_{k=0}^{k} (-1)^{k}\binom{m}{k}{s}^{k}ds\\ =\sum_{k=0}^{n} (-1)^{k}\binom{m}{k}\int_{0}^{1}{s}^{k+n}ds\\ =\sum_{k=0}^{m} (-1)^{k}\binom{m}{k} \frac{1}{k+n+1}　 $$