Asymptote for $\frac{\left(\Gamma(1+1/t)\right)^2}{\Gamma(1+2/t)}$

Question

I was trying to solve the integral $$I(r, t) = \int_0^r (r^t-x^t)^{1/t}dx$$

I made the substitution $u = x/r$ to get $$r^2\int_0^1(1-u^t)^{1/t}du$$

From here, I used WolframAlpha to get that $$I = r^2\cdot\frac{\left(\Gamma(1+1/t)\right)^2}{\Gamma(1+2/t)}$$

I want to know an infinite summation for $f(t)=I/r^2$, or at least tight bounds. I tried using a Maclaurin series, but each derivative is undefined at $x=0$. I know for a fact that a loose upper bound is $y=1$, while a loose lower bound is $y=0$. However, I am looking for more tight bounds.

Any help in finding an infinite sum or tighter bounds for $f(t)$ would be appreciated.

Answer 1

Considering $$y=\frac{\Gamma \left(1+\frac{1}{t}\right)^2}{\Gamma \left(1+\frac{2}{t}\right)}\implies \log(y)=2 \log\left(\Gamma \left(1+\frac{1}{t}\right)\right)-\log\left(\Gamma \left(1+\frac{2}{t}\right)\right)$$

Now, by Taylor $$\log\left(\Gamma \left(1+\epsilon\right)\right)=\sum_{n=1}^\infty \frac{\psi ^{(n-1)}(1)}{n (n-1)!}\epsilon^n$$ Truncated for a few terms $$\log\left(\Gamma \left(1+\epsilon\right)\right)=-\gamma \epsilon +\frac{\pi ^2 \epsilon ^2}{12}+\frac{\epsilon ^3 \psi ^{(2)}(1)}{6}+\frac{\pi ^4 \epsilon ^4}{360}+O\left(\epsilon ^5\right)$$ Then $$\log(y)=-\sum_{n=1}^\infty \frac{\left(2^n-2\right) \psi ^{(n-1)}(1)}{n!}\frac 1{t^n}$$ Truncated for a few terms $$\log(y)=-\frac{\pi ^2}{6 t^2}+\frac{2 \zeta (3)}{t^3}-\frac{7 \pi ^4}{180 t^4}+O\left(\frac{1}{t^5}\right)$$ $$y=e^{\log(y)}=1-\frac{\pi ^2}{6 t^2}+\frac{2 \zeta (3)}{t^3}-\frac{\pi ^4}{40 t^4}-\frac{\pi ^2 \zeta (3)}{3 t^5}+O\left(\frac{1}{t^6}\right)$$

The relative error is $< 0.1$% as soon as $t >6$ and $< 0.01$% as soon as $t >10$.

Edit

You could have a better more compact approximation using a Padé approximant. It would be $$y\sim \frac{1-\frac{6 \psi ^{(2)}(1)}{\pi ^2 t}+\frac{\frac{36 \psi ^{(2)}(1)^2}{\pi^4}-\frac{19 \pi ^2}{60}}{t^2} } {1-\frac{6 \psi ^{(2)}(1)}{\pi ^2 t}+\frac{\frac{36 \psi ^{(2)}(1)^2}{\pi ^4}-\frac{3 \pi ^2}{20}}{t^2} }$$

Edit

We could make the problem more general considering $$y=\frac{\Gamma \left(1+\frac{1}{t}\right)^a}{\Gamma \left(1+\frac{a}{t}\right)}$$ and get $$y=1-\frac{\pi ^2 a(a-1) }{12 t^2}+\frac{a \left(a^2-1\right) \zeta (3)}{3 t^3}+\frac{\pi ^4 a(a-1) (a^2-9 a-4)}{1440 t^4}+O\left(\frac{1}{t^5}\right)$$

Answer 2

regularly we always have expansion over $\ln\Gamma(z)$ instead, here you take $2\ln(1+z)-\ln(1+2z)$

then for small $z$ you have ($\zeta(k)$ refers to Riemann Zeta function)

$$\ln\Gamma(1+z)=-\gamma z+\sum_{k=2}^{\infty}\frac{\zeta(k)}{k}(-z)^k$$

for large $z$ you have $\ln\Gamma(1+z)=\ln z+\ln\Gamma(z)$, where

$$\ln\Gamma(z)=z\ln z-z-\frac1{2}\ln z+\frac1{2}\ln(2\pi)+\frac1{12z}-\frac1{360z^3}+O(z^5)$$

the rests are some simple calculation.

Answer 3

We have the following inequalities for the Gamma function:

$$\sqrt{2\pi x} \left(\frac{x}{e}\right)^x < \Gamma(1+x) < e\sqrt{x} \left(\frac{x}{e}\right)^x$$

for all $x > 1$. For $x\leq 1$, only the left side inequality is true. Focusing on the case $t<1$ we have that

$$\frac{\left(\Gamma\left(1+\frac{1}{t}\right)\right)^2}{\Gamma\left(1+\frac{2}{t}\right)} < \frac{e^2}{2\sqrt{\pi}}\cdot \frac{4^{-\frac{1}{t}}}{\sqrt{t}}$$

Asymptotically though, Stirling's approximation tells us that

$$\frac{\left(\Gamma\left(1+\frac{1}{t}\right)\right)^2}{\Gamma\left(1+\frac{2}{t}\right)} \sim 4^{-\frac{1}{t}}\sqrt{\frac{\pi}{t}}$$

For the case $t \geq 1$, we can use the following inequalities

$$x^{\frac{x}{2}} \leq \Gamma(1+x) \leq x^{-\frac{x}{2}}$$

for all $x\leq 1$ to get the bounds

$$\left(\frac{t^2}{2}\right)^{-\frac{1}{t}} < \frac{\left(\Gamma\left(1+\frac{1}{t}\right)\right)^2}{\Gamma\left(1+\frac{2}{t}\right)} < \left(\frac{t^2}{2}\right)^{\frac{1}{t}}$$

Answer 4

Starting from $$ J = I/r^{\,2} = {{\left( {\Gamma \left( {1 + 1/t} \right)} \right)^{\,2} } \over {\Gamma \left( {1 + 2/t} \right)}} $$

and applying the functional identity for Gamma we get $$ \Gamma \left( {1 + z} \right) = z\,\Gamma \left( z \right)\quad \Rightarrow \quad J = {1 \over {2t}}{{\left( {\Gamma \left( {1/t} \right)} \right)^{\,2} } \over {\Gamma \left( {2/t} \right)}} $$

Now we can apply the duplication formula for Gamma $$ \eqalign{ & \Gamma \left( {2\,z} \right) = {{2^{\,2\,z - 1} } \over {\sqrt \pi }}\Gamma \left( z \right)\Gamma \left( {z + 1/2} \right)\quad \Rightarrow \cr & \Rightarrow \quad J = {1 \over {2t}}{{\left( {\Gamma \left( {1/t} \right)} \right)^{\,2} } \over {{{2^{\,2\,/t - 1} } \over {\sqrt \pi }}\Gamma \left( {1/t} \right)\Gamma \left( {1/t + 1/2} \right)}} = \cr & = {{\sqrt \pi } \over {t2^{\,2\,/t} }}{{\Gamma \left( {1/t} \right)} \over {\Gamma \left( {1/t + 1/2} \right)}} = {{\sqrt \pi } \over {t2^{\,2\,/t} }}{1 \over { \left( {1/t} \right)^{\,\overline {\,1/2\,} } }} = \cr & = {1 \over {t2^{\,2\,/t} }}{{\Gamma \left( {1/2} \right)\Gamma \left( {1/t} \right)} \over {\Gamma \left( {1/t + 1/2} \right)}} = {1 \over {t2^{\,2\,/t} }}{\rm B}\left( {1/2,\;1/t} \right) = \cr & = {{\sqrt \pi \,\Gamma \left( {3/2} \right)} \over {2^{\,2\,/t} }}{{\Gamma \left( {1/t + 1} \right)} \over {\Gamma \left( {1/t + 1/2} \right)\Gamma \left( {3/2} \right)}} = {{\,\pi } \over {2^{\,2\,/t + 1} }} \binom{1/t}{1/2} \cr } $$ so that you have here an expression for $J$ in terms of:
- rising Factorial;
- Beta function;
- Binomial.
which are considered to be closed expressions.