Question Number 60797 by arcana last updated on 25/May/19
$$\int\frac{{e}^{{w}} }{{w}^{{n}+\mathrm{1}} }{dw},\:{n}\in\mathbb{N} \\ $$
Commented by MJS last updated on 26/May/19
![this reminds me of Γ(x)=∫_0 ^∞ e^(−t) t^(x−1) dt ∫e^w w^(−n−1) dw= [v=w^(−n) → dw=−(w^(n+1) /n)dv] =−(1/n)∫e^(1/(v)^(1/n) ) dv this is a special integral called incomplete gamma function. I found this online ∫e^(1/(v)^(1/n) ) dv=(−1)^n nΓ(−n, −(1/( (v)^(1/n) ))) ⇒ −(1/n)∫e^(1/(v)^(1/n) ) dv=−(−1)^n Γ(−n, −(1/( (v)^(1/n) )))= =−(−1)^n Γ(−n, −w)+C sorry cannot further explain...](https://www.tinkutara.com/question/Q60850.png)
$$\mathrm{this}\:\mathrm{reminds}\:\mathrm{me}\:\mathrm{of}\:\Gamma\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\mathrm{e}^{−{t}} {t}^{{x}−\mathrm{1}} {dt} \\ $$$$\int\mathrm{e}^{{w}} {w}^{−{n}−\mathrm{1}} {dw}= \\ $$$$\:\:\:\:\:\left[{v}={w}^{−{n}} \:\rightarrow\:{dw}=−\frac{{w}^{{n}+\mathrm{1}} }{{n}}{dv}\right] \\ $$$$=−\frac{\mathrm{1}}{{n}}\int\mathrm{e}^{\mathrm{1}/\sqrt[{{n}}]{{v}}} {dv} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{special}\:\mathrm{integral}\:\mathrm{called}\:{incomplete} \\ $$$${gamma}\:{function}.\:\mathrm{I}\:\mathrm{found}\:\mathrm{this}\:\mathrm{online} \\ $$$$\int\mathrm{e}^{\mathrm{1}/\sqrt[{{n}}]{{v}}} {dv}=\left(−\mathrm{1}\right)^{{n}} {n}\Gamma\left(−{n},\:−\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{v}}}\right) \\ $$$$\Rightarrow\:−\frac{\mathrm{1}}{{n}}\int\mathrm{e}^{\mathrm{1}/\sqrt[{{n}}]{{v}}} {dv}=−\left(−\mathrm{1}\right)^{{n}} \Gamma\left(−{n},\:−\frac{\mathrm{1}}{\:\sqrt[{{n}}]{{v}}}\right)= \\ $$$$=−\left(−\mathrm{1}\right)^{{n}} \Gamma\left(−{n},\:−{w}\right)+{C} \\ $$$$\mathrm{sorry}\:\mathrm{cannot}\:\mathrm{further}\:\mathrm{explain}… \\ $$
Commented by arcana last updated on 26/May/19
$$\mathrm{muchas}\:\mathrm{gracias},\mathrm{I}\:\mathrm{need}\:\mathrm{this}\:\mathrm{but}\:\mathrm{now}\: \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{some}\:\mathrm{idea}\::\mathrm{D} \\ $$