MHB How do I find the integral of t^{x-1} e^{-t} from 0 to 1?

[Alone]

I am reading the red book on special functions of Andrews, and he writes there that:

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

And I don't see how to arrive at this identity, I guess he expands t in the integrand but my memory is rusty as to this series expansion.

Thanks in advance.

 

[alyafey22]

Re: estimate of an integral.

Here is a hint, use the maclurain representation of $$e^{-t}$$ ...

 

[Chris L T521]

Re: estimate of an integral.

I am reading the red book on special functions of Andrews, and he writes there that:

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

And I don't see how to arrive at this identity, I guess he expands t in the integrand but my memory is rusty as to this series expansion.

Thanks in advance.

Edit: I was ninja'd by ZaidAlyafey, so I'll put the work in spoilers instead of deleting it all... xD

The gist of the argument is this -- you express $e^{-t}$ as a Taylor series: $\displaystyle e^{-t}=\sum_{n=0}^{\infty}\frac{(-1)^nt^n}{n!}$

Spoiler

So it now follows that
\[\int_0^1t^{x-1}e^{-t}\,dt =\int_0^{\infty}t^{x-1} \sum_{n=0}^{\infty}\frac{(-1)^n t^n}{n!}\,dt = \int_0^1\sum_{n=0}^{\infty} (-1)^n\frac{t^{x+n-1}}{n!} \,dt\]
We now integrate termwise to get
\[\sum_{n=0}^{\infty}(-1)^n \left(\int_0^1 \frac{t^{x+n-1}}{n!} \,dt\right)= \sum_{n=0}^{\infty} (-1)^n\left[\frac{t^{x+n}}{(x+n) \cdot n!}\right]_0^1 =\sum_{n=0}^{\infty} \frac{(-1)^n}{(x+n)\cdot n!}\]
which is what was to be shown.

I hope this makes sense!

 

[Alone]

Re: estimate of an integral.

Thanks, what an idiot I am not to see this straight away... I am getting old. :-)

 

[alyafey22]

Re: estimate of an integral.

I am reading the red book on special functions of Andrews

Really , interesting , may I ask what is the purpose ?

 

[chisigma]

Re: estimate of an integral.

I am reading the red book on special functions of Andrews, and he writes there that:

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

And I don't see how to arrive at this identity, I guess he expands t in the integrand but my memory is rusty as to this series expansion.

Thanks in advance.

You have...

$\displaystyle t^{x-1} e^{-t} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!}\ t^{x-1+n}$ (1)

... so that is...

$\displaystyle \int_{0}^{1} t^{x-1} e^{-t} dt = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!}\ \int_{0}^{1} t^{x-1+n}\ dt = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(x+n)\ n!}$ (2)

Kind regards

$\chi$ $\sigma$