انتگرالها را جدا حساب میکنیم:
$1) \sqrt{e^x}=u \Rightarrow e^x=u^2 \Rightarrow e^xdx=2udu \Rightarrow dx= \frac{2}{u} du$
$ \Rightarrow \int _0^ee^ \sqrt{e^x}dx=2 \int_1^b \frac{e^u}{u} du=2 \int _{- \infty }^b \frac{e^u}{u} du-2 \int _{- \infty }^1 \frac{e^u}{u} du=2(Ei(b)-Ei(1))$
$2)Ln(Lnx)=v \Rightarrow Lnx=e^v \Rightarrow x=e^{e^v} \Rightarrow dx=e^ve^{e^v}dv$
$ \Rightarrow \int_e^bLn(Lnx)dx= \int _0^{0.5}ve^ve^{e^v}dv= \int_0^{0.5}vd(e^{e^v})$
$=[ve^{e^v}]_0^{0.5}- \int _0^{0.5}e^{e^v}dv= \frac{b}{2}-\int _0^{0.5}e^{e^v}dv$
حالا قرار دهید:
$e^v=w \Rightarrow e^vdv=dw \Rightarrow \int _0^{0.5}e^{e^v}dv= \int _1^{ \sqrt{e} } \frac{e^w}{w} dw$
$= \int _{- \infty }^{ \sqrt{e} } \frac{e^w}{w} dw- \int _{- \infty }^1\frac{e^w}{w} dw=Ei( \sqrt{e} )-Ei(1)$
$ \Rightarrow \int_e^bLn(Lnx)dx= \frac{b}{2} -Ei( \sqrt{e} )+Ei(1)$
$ \Box $
