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A few days ago, I posted a partial solution to an old question re. showing

$$\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4} \, dx = \frac\pi4$$

by converting $\frac1{1+x^4}$ to a Taylor series and I ended up recovering a sum involving a ratio of Gamma functions. The missing piece is to show

$$\sum_{k=0}^\infty (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)} = 2\sqrt\pi$$

Similar resolved questions seem to suggest the duplication or multiplication formulae would be of some help and/or the identities showcased here, but I'm not sure how to align the sum I have to match the forms given in the identity.

I'm most interested in a hint - what manipulations should I consider to get a form that more closely resembles the known results above? Any other methods are acceptable, but I'm specifically trying to compute this sum in order to evaluate the aforementioned integral.


By converting Gammas to factorials to binomial coefficients, I got

$$\begin{align*} \sum_{k=0}^\infty (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)} &= \left(-\frac32\right)! \sum_{k=0}^\infty (-1)^k \frac{\left(k-\frac34\right)!}{\left(k+\frac34\right)!\left(-\frac32\right)!} \\[1ex] &= \Gamma\left(-\frac12\right) \sum_{k=0}^\infty (-1)^k \binom{k-\frac34}{k+\frac34} \\[1ex] &= 2\sqrt\pi \sum_{k=0}^\infty \binom{\frac{8k-11}4}{\frac{8k-1}4} \end{align*}$$

where in the last equality, I applied Pascal's rule $\binom nk=\binom{n-1}k+\binom{n-1}{k-1}$ (though I'm not 100% sure it holds for $n,k\in\Bbb Q$) to eliminate the positive terms in the series. So the new missing piece is

$$\sum_{k=1}^\infty \binom{\frac{8k-11}4}{\frac{8k-1}4} = 1$$

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  • $\begingroup$ It is equivalent to show: $$\sum_{k=0}^\infty(-1)^k\mathfrak{B}(k+1/4,3/2)=\pi$$Although this possibly leads to the original integral $\endgroup$
    – FShrike
    Aug 3, 2022 at 20:19
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    $\begingroup$ Why are you banning backtracking through the link? Using those methods might be the most natural way to solve it had you not even seen that question before. It seems arbitrary. $\endgroup$
    – Jacob
    Aug 3, 2022 at 23:25
  • $\begingroup$ The sum is also $$\sum_{k=0}^\infty \frac{(-1)^k}{\left(k+\frac14\right)_\frac32}$$ with the pochhammer symbol $\endgroup$ Aug 4, 2022 at 3:15
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    $\begingroup$ @Jacob I don't see the harm in seeking an alternate solution, regardless of how unnatural it may appear. People on this site ask about computing limits and integrals without resorting to specific methods like l'Hopital's rule or complex analysis all the time. $\endgroup$
    – user170231
    Aug 4, 2022 at 3:33
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    $\begingroup$ @user170231 I believe I see what you mean. To be clear, I don't think it is wrong to look for solutions that don't use certain methods. I just think "don't use any of the methods in this post" is far more arbitrary than "don't use [specific methods]" because what people choose to post is a variable. I think it would have helped if you were more direct in saying "I want to prove this sum equality to prove this integral equality - that is, any method that proves the integral equality first is useless to me," which I think is a completely valid take. $\endgroup$
    – Jacob
    Aug 4, 2022 at 20:22

4 Answers 4

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Not sure if this counts as "backtracking" since your answer in the first link already mentions this equation

$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}=\frac 2{\sqrt\pi}\sum\limits_{k=0}^{+\infty}(-1)^k\int\limits_0^1t^{k-3/4}(1-t)^{1/2}\,\mathrm dt$$

But it is easier to evaluate the sum by employing the Beta function and integrating the sum of the expression inside the integral. Swapping the order of the integral and summation before applying the geometric series leads to

$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}=\frac 2{\sqrt\pi}\int\limits_0^1\frac {\sqrt{1-t}}{t^{3/4}(1+t)}\,\mathrm dt$$

This resulting integral might seem more convoluted, however the simple substitution $u=\frac {1-t}{1+t}$ reduces it into another form of the beta function which can then be further simplified using Euler's Reflection formula.

\begin{align*} \int\limits_0^1\frac {\sqrt{1-t}}{t^{3/4}(1+t)}\,\mathrm dt & =\sqrt2\int\limits_0^1u^{1/2}\left(1-u^2\right)^{-3/4}\,\mathrm du\\ & =\frac 1{\sqrt2}\int\limits_0^1u^{-1/4}(1-u)^{-3/4}\,\mathrm du\\ & =\frac 1{\sqrt2}\Gamma\left(\frac 14\right)\Gamma\left(\frac 34\right)\\ & =\pi \end{align*}

This proves the result.

$$\sum\limits_{k=0}^{+\infty}(-1)^k\frac {\Gamma\left(k+\frac 14\right)}{\Gamma\left(k+\frac 74\right)}\color{blue}{=2\sqrt\pi}$$

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Compute the partial sum $$S_p=\sum_{k=0}^p (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)} $$ and using the Gaussian hypergeometric function $$S_p=\frac{8 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{3 \Gamma \left(\frac{3}{4}\right)}+(-1)^p\frac{ \, _2F_1\left(1,p+\frac{5}{4};p+\frac{11}{4};-1\right) \Gamma \left(p+\frac{5}{4} \right)}{\Gamma \left(p+\frac{11}{4} \right)}$$ $$S_p=2 \sqrt{\pi }+(-1)^p \, _2\tilde{F}_1\left(1,p+\frac{5}{4};p+\frac{11}{4};-1\right) \Gamma \left(p+\frac{5}{4}\right)$$ For $p=10$, the second term is,numerically, $0.013661$. For $p=100$, it is $0.000493$.

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Here is another evaluation using the hypergeometric function:

$$\sum_{k=1}^\infty\frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)}x^k=\frac{4\Gamma\left(\frac14\right)\,_2\text F_1\left(1,\frac14;\frac74;x\right)}{3\Gamma\left(\frac34\right)}$$

Now use this incomplete beta function identity:

$$\frac{4\Gamma\left(\frac14\right)\,_2\text F_1\left(1,\frac14;\frac74;-1\right)}{3\Gamma\left(\frac34\right)} =-\frac{(1+i)\Gamma^2\left(\frac14\right)\text B_{-1}\left(\frac34,-\frac12\right)}{\sqrt2\pi}$$

Now convert back to a hypergeometric function like so:

$$-\frac{(1+i)\Gamma^2\left(\frac14\right)\text B_{-1}\left(\frac34,-\frac12\right)}{\sqrt2\pi}=-\frac{4(1+i)(-1)^\frac34\Gamma^2\left(\frac14\right)}{3\sqrt2\pi}\,_2\text F_1\left(\frac32,\frac34; \frac74;-1\right) $$

to finish off use this identity:

$$-\frac{4(1+i)(-1)^\frac34\Gamma^2\left(\frac14\right)}{3\sqrt2\pi}\,_2\text F_1\left(\frac32,\frac34; \frac74;-1\right)=\frac{4\Gamma^2\left(\frac14\right)\Gamma\left(\frac34\right)\Gamma\left(\frac74\right)}{3\pi^\frac32}=2\sqrt\pi$$

No integration needed

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Let $$S=\sum_{k=0}^\infty (-1)^k \frac{\Gamma\left(k+\frac14\right)}{\Gamma\left(k+\frac74\right)}$$ Using beta-integral and gamma functions we can write $$S=\frac{2}{\Gamma(3/2)}\sum_{0}^{\infty} \int_{0}^{\pi/2} \sin^{2k-1/2}x~ \cos ^2 x~ dx.$$ Using IGP $$\implies S=\frac{2}{\Gamma(3/2)}\int_{0}^{\pi/2} \frac{\cos^2x}{\sqrt{\sin x} (1+\sin^2 x)} dx$$ Let $\sin x =t^2$, we get $$S=\frac{4}{\sqrt{\pi}}\int_{0}^{1}\frac{2\sqrt{1-t^4}}{1+t^4} dt$$ This integral has been evaluated in Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$, we get $$S=\frac{8}{\sqrt{\pi}} \frac{\pi}{4}=2 \sqrt{\pi}$$

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    $\begingroup$ I believe OP is trying to prove the integral highlighted by evaluating the infinite sum, not use the integral to aid in the evaluation of the sum $\endgroup$
    – Frank W
    Aug 4, 2022 at 7:35
  • $\begingroup$ I have edited it now. $\endgroup$
    – Z Ahmed
    Aug 4, 2022 at 9:10
  • $\begingroup$ What do you mean by IGP? $\endgroup$
    – user170231
    Aug 4, 2022 at 15:34
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    $\begingroup$ infinite Geometric progression. $\endgroup$
    – Z Ahmed
    Aug 4, 2022 at 16:18

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