- Thread starter
- #1

#### Also sprach Zarathustra

##### Member

- Jan 31, 2012

- 54

$$ \int_0^1 x^{-x}dx=\sum_{n=1}^{\infty}f_n $$Find $f_n$.

- Thread starter Also sprach Zarathustra
- Start date

- Thread starter
- #1

- Jan 31, 2012

- 54

$$ \int_0^1 x^{-x}dx=\sum_{n=1}^{\infty}f_n $$Find $f_n$.

- Thread starter
- #3

- 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.

- 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 \(\displaystyle f_n = \frac{1}{n^n} \)

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:

- Feb 6, 2012

- 20

How do you prove this statement?4. Note That: \( \displaystyle \int_{0}^{1} x^n \ln^m(x) dx=\frac{(-1)^m m!}{(n+1)^{m+1}}\)

- Jan 27, 2012

- 95

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

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

- Jan 31, 2012

- 253

i can't understand 3. line1. 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 \(\displaystyle f_n = \frac{1}{n^n} \)

- Feb 9, 2012

- 118

A power series expansion was done.

- Jan 27, 2012

- 95

Taylor Series,i can't understand 3. line

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

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