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I am trying to extend the Sophomore's Dream by proving the formula:

$$\int _0^1\limits x^{a-x}\text d x=\sum _{n=1}^\infty \limits \frac{1}{(a+n)^n}$$

I have tried induction, integration by parts, and using the same proof as here and substituting $a-x$, but haven't had much luck.

Thanks in advance!

Andrew Dynneson
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3 Answers3

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$$ x^{a-x}=x^a e^{-x\ln x}=x^a\left(\sum_{n=0}^{\infty}(-1)^n\frac{x^n\ln^n x}{n!}\right)=\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+a}\ln^n x}{n!}. $$ Intergrate by parts $$ \int_0^1 (-1)^n\frac{x^{n+a}\ln^n x}{n!}\mathrm{d}x=(-1)^{n}\frac{x^{n+a+1}\ln^n x}{(n+a+1)n!}|_{x=0}^{x=1}+(-1)^{n-1}\int_0^1 \frac{x^{n+a}\ln^{n-1}x}{(n+a+1)(n-1)!}\mathrm{d}x $$ $$ =\cdots=\frac{1}{(n+a+1)^{n+1}}. $$ Hence we have $$ \int _0^1\limits x^{a-x}\text d x=\sum _{n=1}^\infty \limits \frac{1}{(a+n)^n}. $$

yahoo
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  • It checks out, thank you. I think there is a sign-error at $-(-1)^{n-1}$ should be $+(-1)^{n-1}$. It was easy to show $\lim _{x \rightarrow 0}\limits x^{n + a + 1} \ln ^n x =0$. – Andrew Dynneson Jan 10 '18 at 05:51
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$$\int _0^1\limits x^{a-x}\text d x=\int_0^1e^{(a-x)\ln(x)}dx=\int_0^1\sum_{n=0}^\infty\frac{((a-x)\ln(x))^n}{n!}dx=\sum_{n=0}^\infty\frac{1}{n!}\int_0^1((a-x)\ln(x))^ndx$$ Let $u=-\ln(x)$ and $-e^udu=dx$

So $$\sum_{n=0}^\infty\frac{1}{n!}\int_0^1((a-x)\ln(x))^ndx=\sum_{n=0}^\infty\frac{1}{n!}\int_0^\infty(e^{-u}-a)^nu^ne^udu$$

Now recall that $n!=\int_0^\infty u^n e^{-u}du$

Can you continue from here?

aleden
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1

Although, the answer is already published, I would like to leave a comment here. There is a nice “umbral calculus” method, which is, of course, the same, but is pretty intuitively clear.

Consider the Taylor formulas:

$$ f(\partial/\partial x) e^{Ax}=f(A)e^{Ax}, ~~e^{A\partial/\partial x}f(x)=f(x+A); $$

the identity

$$f(\partial/\partial x)g(x)|_{x=0}=\sum_{n=0}^\infty \frac{f^{(n)}(0)g^{(n)}(0)}{n!}=g(\partial/\partial x)f(x)|_{x=0} ;$$

and the equality $\frac{1}{x}=\int_0^{\infty} e^{-xt}dt$.

Then:

\begin{align*} \sum_{n=1}^\infty \frac{x^n}{(n+a)^n}&=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n (\partial/\partial y)^{n-1}}{(n-1)!}~ \frac{1}{n+a+y}\Big|_{y=0}=\\ ~\\ &=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n (\partial/\partial y)^{n-1}e^{(n-1+a)\partial/\partial y}}{(n-1)!}~ \frac{1}{1+y}\Big|_{y=0}=\\ ~\\ &=x\exp\left(a\frac{\partial}{\partial y}-x\frac{\partial}{\partial y}e^{\partial/\partial y}\right)~\frac{1}{1+y}\Big|_{y=0}=\\ ~\\ &=\frac{1}{1+\frac{\partial}{\partial y}}~x\exp\left(ay-xye^{y}\right)\Big|_{y=0}=\\ ~\\ &=\int_{0}^{\infty} e^{-t-t\partial/\partial y}dt ~~~ x\exp(ay-xye^y)\Big|_{y=0}=\\ ~\\ &=x\int_{0}^\infty e^{-t}\exp(a(y-t)-x(y-t)e^{y-t})dt\Big|_{y=0}=\\ ~\\ &=x\int_{0}^\infty e^{-t(1+a)+xte^{-t}}dt=\\ ~\\ \tag*{qed}&=x\int_{0}^{1} t^{a-xt}dt \end{align*}

Danil Krotkov
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