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Integral of x^{-x}

Jan 31, 2012
54
$$ \int_0^1 x^{-x}dx=\sum_{n=1}^{\infty}f_n $$Find $f_n$.
 

Markov

Member
Feb 1, 2012
149
Very easy, it's well known, I think it doesn't even count as a "challenge," and why is this, because you express $x^{-x}$ as $e^{-x\ln x}$ and use power series, that's all.
 
Jan 31, 2012
54
Very easy, it's well known, I think it doesn't even count as a "challenge," and why is this, because you express $x^{-x}$ as $e^{-x\ln x}$ and use power series, that's all.

Yes. You are right it is not a "challenge"! This is what happens when you have nothing to do on 03:00 am. :(
 

sbhatnagar

Active member
Jan 27, 2012
95
1. Let \( \displaystyle I=\int_{0}^{1}x^{-x} dx \)

2. \(\displaystyle x^{-x}= e^{-x \ln(x)}= 1-\frac{x\ln(x)}{1!}+\frac{x^2\ln^2(x)}{2!}-\frac{x^3\ln^3(x)}{3!}+\cdots\)

3. \( \displaystyle \int_{0}^{1}x^{-x} dx= \int_{0}^{1}e^{-x \ln(x)} dx = \int_{0}^{1} \Big( 1-\frac{x\ln(x)}{1!}+\frac{x^2\ln^2(x)}{2!}-\frac{x^3\ln^3(x)}{3!}+\cdots \Big)dx\)

4. Note That: \( \displaystyle \int_{0}^{1} x^n \ln^m(x) dx=\frac{(-1)^m m!}{(n+1)^{m+1}}\)

5. Integrating term wise by using the above formula we obtain:\( \displaystyle I=1 +\frac{1}{2^2}+\frac{1}{3^3}+\cdots =\sum_{n=1}^{\infty}\frac{1}{n^n} \)

So \(\displaystyle f_n = \frac{1}{n^n} \)
 
Last edited:

DrunkenOldFool

New member
Feb 6, 2012
20

sbhatnagar

Active member
Jan 27, 2012
95
How do you prove this statement?
\( \displaystyle \int_{0}^{1}x^n \ln^m(x) dx =\int_{0}^{\infty}e^{-x(n+1)} (-x)^m dx=(-1)^m \int_{0}^{\infty} e^{-x(n+1)} x^m=\frac{(-1)^m \Gamma(m+1)}{(n+1)^{m+1}}=\frac{(-1)^m m!}{(n+1)^{m+1}}\)
 

Markov

Member
Feb 1, 2012
149
3. \( \displaystyle \int_{0}^{1}x^{-x} dx= \int_{0}^{1}e^{-x \ln(x)} dx = \int_{0}^{1} \Big( 1-\frac{x\ln(x)}{1!}+\frac{x^2\ln^2(x)}{2!}-\frac{x^3\ln^3(x)}{3!}+\cdots \Big)dx\)
This step is important, we can swap sum an integral because of the Monotone Convergence Theorem, since the terms of the series are non-negative.
 

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
Why would we need to invoke a convergence theorem from Lebesgue integration when the exponential series converges uniformly?
 

Markov

Member
Feb 1, 2012
149
It doesn't matter, you're just justifying it.
 

oasi

New member
Mar 14, 2012
14
1. Let \( \displaystyle I=\int_{0}^{1}x^{-x} dx \)

2. \(\displaystyle x^{-x}= e^{-x \ln(x)}= 1-\frac{x\ln(x)}{1!}+\frac{x^2\ln^2(x)}{2!}-\frac{x^3\ln^3(x)}{3!}+\cdots\)

3. \( \displaystyle \int_{0}^{1}x^{-x} dx= \int_{0}^{1}e^{-x \ln(x)} dx = \int_{0}^{1} \Big( 1-\frac{x\ln(x)}{1!}+\frac{x^2\ln^2(x)}{2!}-\frac{x^3\ln^3(x)}{3!}+\cdots \Big)dx\)

4. Note That: \( \displaystyle \int_{0}^{1} x^n \ln^m(x) dx=\frac{(-1)^m m!}{(n+1)^{m+1}}\)

5. Integrating term wise by using the above formula we obtain:\( \displaystyle I=1 +\frac{1}{2^2}+\frac{1}{3^3}+\cdots =\sum_{n=1}^{\infty}\frac{1}{n^n} \)

So \(\displaystyle f_n = \frac{1}{n^n} \)
i can't understand 3. line
 

Krizalid

Active member
Feb 9, 2012
118
A power series expansion was done.
 

sbhatnagar

Active member
Jan 27, 2012
95
i can't understand 3. line
Taylor Series,

$\displaystyle e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$

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