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Zanzabum
3 years ago
12

So, my buddy is new to doing Calculus and needs help understanding this equation, it would be very appeciated for some help

Mathematics
1 answer:
mezya [45]3 years ago
7 0

To evaluate the integral, rewrite the integrand as

x^{-x}=e^{\ln x^{-x}}=e^{-x\ln x}

Recall that

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

The leftmost sum is the well-known power series expansion for the function f(x)=e^x. In the rightmost sum, we just replace x with -x\ln x.

This particular power series has a property called "uniform convergence". Roughly speaking, it's a property that says a sequence of functions f_n(x) converges to some limiting function f(x) in the sense that f_n(x) and f_{n+1}(x) get arbitrarily close to one another. If you have an idea of what "convergence" alone means, then you can think of "uniform convergence" as a more powerful form of convergence.

Long story short, this property allows us to interchange the order of summation/integration to write

\displaystyle\int_0^1x^{-x}\,\mathrm dx=\int_0^1\sum_{n=0}^\infty\frac{(-x\ln x)^n}{n!}\,\mathrm dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\int_0^1(x\ln x)^n\,\mathrm dx

The integral can be tackled with a substitution,

x=e^{-u/(n+1)}\implies-(n+1)\ln x=u\implies\mathrm dx=-\dfrac1{n+1}e^{-u/(n+1)}\,\mathrm du

so that the integral is equivalent to

\displaystyle\int_0^1(x\ln x)^n\,\mathrm dx=\int_\infty^0\left(e^{-u/(n+1)}\right)^n\left(-\frac u{n+1}\right)^n\left(-\frac1{n+1}e^{-u/(n+1)}\right)\,\mathrm du

=\displaystyle\frac{(-1)^n}{(n+1)^{n+1}}\int_0^\infty e^{-u}u^n\,\mathrm du

The remaining integral reduces to n!, which you can derive for yourself via integration by parts/power reduction.

So we have

\displaystyle\int_0^1x^{-x}\,\mathrm dx=\sum_{n=0}^\infty\frac{(-1)^n}{n!}\cdot\frac{(-1)^nn!}{(n+1)^{n+1}}=\sum_{n=0}^\infty\frac1{(n+1)^{n+1}}

which is the same as

\displaystyle\sum_{n=1}^\infty\frac1{n^n}=\sum_{n=1}^\infty n^{-n}

and hence the identity.

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