Integral of x^{-x}

Also sprach Zarathustra

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

• sbhatnagar

Markov

Member
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.

Also sprach Zarathustra

Member
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
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:
• Also sprach Zarathustra

DrunkenOldFool

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

sbhatnagar

Active member
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
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
Why would we need to invoke a convergence theorem from Lebesgue integration when the exponential series converges uniformly?

Markov

Member
It doesn't matter, you're just justifying it.

oasi

New member
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
A power series expansion was done.

sbhatnagar

Active member
i can't understand 3. line
Taylor Series,

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

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