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Estimate of an integral

Alan

Member
Jul 21, 2012
58
I am reading the red book on special functions of Andrews, and he writes there that:

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

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.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: estimate of an integral.

Here is a hint, use the maclurain representation of \(\displaystyle e^{-t}\) ...
 

Chris L T521

Well-known member
Staff member
Jan 26, 2012
995
Re: estimate of an integral.

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

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

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

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!
 

Alan

Member
Jul 21, 2012
58
Re: estimate of an integral.

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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

Well-known member
Feb 13, 2012
1,704
Re: estimate of an integral.

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

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

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

$\chi$ $\sigma$