已知 F(x)=∫0x(1+(x−t)+(x−t)22!+…+(x−t)n−1(n−1)!)entdt\displaystyle F(x) = \int_0^x \big(1+(x-t)+\frac{(x-t)^2}{2!}+\ldots +\frac{(x-t)^{n-1}}{(n-1)!} \big) e^{nt} dtF(x)=∫0x(1+(x−t)+2!(x−t)2+…+(n−1)!(x−t)n−1)entdt.
求F(n)(0)=dnF(x)dxn∣x=0F^{(n)}(0)=\frac{d^n F(x)}{d x^n}\vert_{x=0}F(n)(0)=dxndnF(x)∣x=0.
解:作代换x−t=sx-t=sx−t=s, 那么F(x)=enx∫0x(1+s+s22+…+sn−1(n−1)!)e−nsdsF(x)=e^{nx} \int_0^x \big(1+s+\frac{s^2}{2}+\ldots + \frac{s^{n-1}}{(n-1)!} \big) e^{-ns} dsF(x)=enx∫0x(1+s+2s2+…+(n−1)!sn−1)e−nsds.
记En−1(s)=1+s+s22+…+sn−1(n−1)!E_{n-1}(s)=1+s+\frac{s^2}{2}+\ldots + \frac{s^{n-1}}{(n-1)!}En−1(s)=1+s+2s2+…+(n−1)!sn−1,则En−1′(x)=En−2(x)E'_{n-1}(x)=E_{n-2}(x)En−1′(x)=En−2(x) (对xxx求导数)。
计算可知
F′(x)=nenx∫0xEn−1(s)e−nsds+En−1(x)=nF(x)+En−1(x)F'(x)=ne^{nx}\int_0^x E_{n-1}(s) e^{-ns} ds + E_{n-1}(x) = nF(x)+E_{n-1}(x)F′(x)=nenx∫0xEn−1(s)e−nsds+En−1(x)=nF(x)+En−1(x),
F′′(x)=n2enx(∫0x⋯ )+nEn−1(x)+En−2(x)=n(nF(x)+En−1(x))+En−2(x)F''(x)=n^2 e^{nx} (\int_0^x \cdots) + nE_{n-1}(x)+E_{n-2}(x)=n(nF(x)+E_{n-1}(x))+E_{n-2}(x)F′′(x)=n2enx(∫0x⋯)+nEn−1(x)+En−2(x)=n(nF(x)+En−1(x))+En−2(x),
…
F(n)(x)=nnF(x)+nn−1+nn−2E2(x)+…+n1(1+x)+n0F^{(n)}(x)=n^n F(x) + n^{n-1}+n^{n-2}E_2(x)+\ldots + n^1(1+x)+n^0F(n)(x)=nnF(x)+nn−1+nn−2E2(x)+…+n1(1+x)+n0, 于是有
F(n)(0)=nn−1+nn−2+…+n+1F^{(n)}(0)=n^{n-1}+n^{n-2} + \ldots + n + 1F(n)(0)=nn−1+nn−2+…+n+1.
例子:
取n=2n=2n=2, F′(x)=2e2x(∫0x⋯)+(1+x)F'(x)=2e^{2x}(\int_0^x \cdots) + (1+x)F′(x)=2e2x(∫0x⋯)+(1+x),
F′′(x)=22e2x(∫0x⋯ )+2(1+x)+1F''(x)=2^2e^{2x}(\int_0^x \cdots ) + 2(1+x) + 1F′′(x)=22e2x(∫0x⋯)+2(1+x)+1, 故F′′(0)=2+1=3F''(0)=2+1=3F′′(0)=2+1=3.