1. ## Integral

HI,

I must show that: $(\forall n\in \mathbb{N}) \sum_{k=0}^n \frac{(-1)^kx^k}{k!}+(-1)^{n+1}\int_0^{x} \frac{(x-t)^n}{n!}e^{-t} dt= e^{-x}$

I find that the expression is true for n=0; but i don't know how to continue?

2. It's enough to show that

$(-1)^{n+1}\int_{0}^{x}\frac{(x-t)^n}{n!}e^{-t}dt=\sum_{k\,=\,n+1}^{\infty}\frac{(-1)^kx^k}{k!}$.

We have

$(-1)^{n+1}\int_{0}^{x}\frac{(x-t)^n}{n!}e^{-t}dt=\frac{(-1)^{n+1}}{n!}\int_{0}^{x}(x-t)^n \left(1-t+\frac{t^2}{2!}-\frac{t^3}{3!}+\cdots \right)dt$.

Now use integration by parts to show that

$\int_{0}^{x}\frac{t^j}{j!}(x-t)^ndt=\frac{x^{n+j+1}}{(n+1)(n+2) \cdots(n+j+1)}$ for $j=0,1,2,\dots$.

3. Originally Posted by lehder
HI,

I must show that: $(\forall n\in \mathbb{N}) \sum_{k=0}^n \frac{(-1)^kx^k}{k!}+(-1)^{n+1}\int_0^{x} \frac{(x-t)^n}{n!}e^{-t} dt= e^{-x}$

I find that the expression is true for n=0; but i don't know how to continue?

Try induction on n: for n = 0 you already did, so now suppose it's true for all k up to n and try to prove for k = n + 1:

Put $I_n:=\sum_{k=0}^n \frac{(-1)^kx^k}{k!}+(-1)^{n+1}\int_0^{x} \frac{(x-t)^n}{n!}e^{-t} dt$ , so doing integration by parts with $u:=(x-t)^{n+1}\,,\,\,v':=e^{-t}$ , we get:

$I_{n+1}:= \sum_{k=0}^{n+1} \frac{(-1)^kx^k}{k!}+(-1)^{n+2}\int_0^{x} \frac{(x-t)^{n+1}}{(n+1)!}e^{-t} dt$ $=\frac{(-1)^{n+1}x^{n+1}}{(n+1)!}+\sum\limits_{k=0}^n\frac{ (-1)^nx^k}{k!}+$ $\frac{(-1)^{n+2}}{(n+1)!}\left\{\left[-(x-t)^{n+1}e^{-t}\right]_0^x-(n+1)\int\limits_0^x (x-t)^ne^{-t}dt\right\}=$

$=\frac{(-1)^{n+1}x^{n+1}}{(n+1)!}+\sum\limits_{k=0}^n\frac{ (-1)^nx^k}{k!}+\frac{(-1)^{n+2}x^{n+1}}{(n+1)!}+(-1)^{n+3}\int\limits_0^x\frac{(x-t)^n}{n!}e^{-t}\,dt$ .

Now just check that you can cancel out the first and the third summands, and the second and fourth ones are exactly the same as $I_n$ and you're done.

Tonio