In this post I present my argument which proves integral no. 5 under the Radical Integrals section of The Integral Corner. The result in question reads as follows.

Radical Integral 5. Let $P$ and $Q$ real satisfy $0< P\leq Q$. Consider the function

$\displaystyle a(x):=1+e\gamma(x+1,1)$

for $x\geq 0$ where $\gamma(u,v)$ is the incomplete gamma function with integral representation

$\displaystyle \gamma(u,v)=\int_0^{v}t^{u-1}e^{-t}\mathrm{d}t.$

Then,

$\displaystyle \int_P^Q\Lambda(x)\log \sqrt[x+1]{a(x)^{x+2}\sqrt[x+2]{a(x)\sqrt[x+3]{a(x)\sqrt[x+4]{a(x)\sqrt[x+5]{\cdots}}}}}\text{ }\mathrm{d}x=\frac{1}{4e}\log\left(H(P,Q)\right)$

with

$\displaystyle \Lambda(x):=\int_0^{1}t^{x}e^{-t}\log(t)\text{ }\mathrm{d}t$

and

$\displaystyle H(P,Q):=\frac{a(Q)^{2a(Q)^2}}{a(P)^{2a(P)^2}}\exp(a(P)^2-a(Q)^2).$

Proof. If we let

$\displaystyle L(x):=\sqrt[x+1]{a(x)^{x+2}\sqrt[x+2]{a(x)\sqrt[x+3]{a(x)\sqrt[x+4]{a(x)\sqrt[x+5]{\cdots}}}}} \text{ }\text{ }(x\geq 0)$

we have for each $x>0$

$\displaystyle L(x)=a(x)^{1+\frac{1}{x+1}+\frac{1}{(x+1)(x+2)}+\frac{1}{(x+1)(x+2)(x+3)}+\cdots}.$

This can be further simplified to

$\displaystyle L(x)=a(x)^{1+e\gamma(x+1,1)}=a(x)^{a(x)}$

after establishing

$\displaystyle \sum_{k=0}^{\infty}\prod_{j=1}^{k+1}\frac{1}{x+j}=e\gamma(x+1,1)$

for $x\geq 0$. To see this, note that the “lower” incomplete gamma function $\gamma(s,z)$ is a holomorphic function with singularities at points $(s,z)$ where $z=0$ or $s$ is a non-positive integer (check out the Incomplete gamma function Wikipedia page). Moreover, it admits the representation

$\displaystyle \gamma(s,z)=z^s\Gamma(s)e^{-z}\sum_{k=0}^{\infty}\frac{z^k}{\Gamma(s+k+1)}.$

Setting $s=x+1$ and $z=1$, we find for each $x>0$

$\displaystyle \gamma(x+1,1)=e^{-1}\Gamma(x+1)\sum_{k=0}^{\infty}\frac{1}{\Gamma(x+k+2)}=e^{-1}\Gamma(x+1)\left(\frac{1}{\Gamma(x+2)}+\frac{1}{\Gamma(x+3)}+\cdots\right)$

Rearranging, we get

$\displaystyle \sum_{k=0}^{\infty}\frac{\Gamma(x+1)}{\Gamma(x+k+2)}=e\gamma(x+1,1).$

But notice that, formally,

$\prod_{j=1}^{k+1}\frac{1}{x+j}\equiv\frac{\Gamma(x+1)}{\Gamma(x+k+2)}.$

As such, we arrive at the desired identity for the infinite sum, justifying the identity $L(x)=a(x)^{a(x)}$ for $x>0$. Writing out $a$ as

$\displaystyle a(x)=1+e\int_0^{1}t^xe^{-t}\mathrm{d}t \text{ }(x\geq 0)$

we see that $a$ is differentiable over $(0,\infty)$ with derivative given by

$\displaystyle \frac{da}{dx}=e\int_0^1t^xe^{-t}\log(t)\mathrm{d}t=e\Lambda(x).$

Consequently, $a$ is monotone decreasing over $(0,\infty)$. Therefore, our proposed integral for given $0 can be evaluated as follows.

$\displaystyle \int_P^Q\Lambda(x)\log{L(x)}\mathrm{d}x=e^{-1}\int_P^Qa(x)\log(a(x))\frac{da}{dx}\mathrm{d}x=e^{-1}\int_{a(P)}^{a(Q)}v\log(v)\mathrm{d}v=e^{-1}\left[\frac{v^2}{2}\log(v)-\frac{1}{4}v^2\right]_{a(P)}^{a(Q)} =\frac{1}{4e}\left[\log\left((v^{2v^2}\exp(-v^{2})\right)\right]_{a(P)}^{a(Q)}.$

Simplification leads to the stated result:

$\displaystyle \int_P^Q\Lambda(x)\log\sqrt[x+1]{a(x)^{x+2}\sqrt[x+2]{a(x)\sqrt[x+3]{a(x)\sqrt[x+4]{a(x)\sqrt[x+5]{\cdots}}}}}\text{ }\mathrm{d}x=\frac{1}{4e}\log\left(H(P,Q)\right)$

with

$\displaystyle \Lambda(x):=\int_0^{1}t^{x}e^{-t}\log(t)\text{ }\mathrm{d}t$

and

$\displaystyle H(P,Q):=\frac{a(Q)^{2a(Q)^2}}{a(P)^{2a(P)^2}}\exp(a(P)^2-a(Q)^2).$

## One thought on “Radical Integrals: No. 5”

1. Brandon Shane Judnarine says:

Glad to see you’ve been keeping busy in quarantine 🙂

Liked by 1 person