Given $ I_n=\int_{0}^{1}\frac{(1-x)^n}{n!}e^x\,dx $, prove that $ I_n=\frac{1}{(n+1)!}+I_{n+1} $

Question

$$ I_n=\int_{0}^{1}\frac{(1-x)^n}{n!}e^x\,dx $$

Prove that $$ I_n=\frac{1}{(n+1)!}+I_{n+1} $$

I tried integration by parts and still can't prove it, I appreciate any hint/answer.

Answer 1

Following the proof of the recurrence relation by Zacky, we have $$I_{n+1}=I_n-\frac{1}{(n+1)!}$$ $$=I_{n-1}-\frac{1}{n!}-\frac{1}{(n+1)!}$$ $$=I_0 - \sum_{k=1}^{n+1} \frac{1}{k!}$$ We can calculate $I_0$ as follows $$I_0 = \int_0^1 e^x dx=[e^x]_0^1=e-1$$ Therefore the final solution is: $$I_{n+1}=e-1 - \sum_{k=1}^{n+1} \frac{1}{k!}$$ $$=e-\sum_{k=0}^{n+1} \frac{1}{k!}$$ or equivalently; $$I_n=e-\sum_{k=0}^{n} \frac{1}{k!}$$

Answer 2

I used $n!$ as in the original image.$$I_n=\int_{0}^{1}\frac{(1-x)^n}{n!}e^x\,dx=\frac{1}{n!}\int_0^1 \left(-\frac{(1-x)^{n+1}}{n+1}\right)'e^x\,dx$$ $$=\underbrace{-\frac{1}{n!}\frac{(1-x)^{n+1}}{n+1} e^x\bigg|_0^1}_{=0-\left(-\large\frac{1}{(n+1)!}\right)} +\int_0^1 \frac{(1-x)^{n+1}}{(n+1)!}e^xdx $$ $$\Rightarrow I_n=\frac{1}{\color{}{(n+1)!}} +I_{n+1}$$