Prove that frac1sin x=frac1x+2x sum from k=1 to infinity
Prove that $\displaystyle\frac1{\sin x}=\frac1x+2x\sum_{k=1}^{\infty}\frac{(-1)^k}{x^2-k^2\pi^2}$. From this, conclude that $\Gamma(x)\Gamma(1-x)=\dfrac\pi{\sin\pi x}$.
Explanation of Method
The cosecant's poles at $x=k\pi$ carry alternating residues $(-1)^k$ (from $1/\cos(k\pi)$), giving its Mittag-Leffler expansion directly. For the reflection formula, the cleanest route uses the Weierstrass canonical product $1/\Gamma(x)=xe^{\gamma x}\prod(1+x/n)e^{-x/n}$: multiplying $\Gamma(x)$ by $\Gamma(-x)$ makes the Euler–Mascheroni terms cancel exactly, leaving a clean product that matches Euler's sine product.
- Derive the cosecant expansion via residues at $x=k\pi$.
- Multiply the Weierstrass products for $\Gamma(x)$ and $\Gamma(-x)$, canceling the $e^{\pm\gamma x}$ factors.
- Identify the resulting product with $\sin(\pi x)/(\pi x)$ and use $\Gamma(-x)=-\Gamma(1-x)/x$ to conclude.
Step 1 · The cosecant expansion
$1/\sin x$ has simple poles at $x=k\pi$ ($k\in\mathbb Z$) with residue $\lim_{x\to k\pi}(x-k\pi)/\sin x=1/\cos(k\pi)=(-1)^k$. The Mittag-Leffler expansion gives $$ \frac1{\sin x} = \frac1x + \sum_{k=1}^\infty(-1)^k\left[\frac1{x-k\pi}+\frac1{x+k\pi}\right] = \frac1x+\sum_{k=1}^\infty(-1)^k\frac{2x}{x^2-k^2\pi^2}. $$ $$ \boxed{\frac1{\sin x} = \frac1x+2x\sum_{k=1}^{\infty}\frac{(-1)^k}{x^2-k^2\pi^2}} $$ (Numerical check at $x=0.7$: partial sum to $10^5$ terms gives $\approx1.552270$, matching $1/\sin(0.7)\approx1.552270$ — the alternating sign gives noticeably faster convergence than Problem 397's cotangent expansion.)
Step 2 · Multiplying Weierstrass products for $\Gamma(x)$ and $\Gamma(-x)$
The canonical Weierstrass product is $\dfrac1{\Gamma(x)}=xe^{\gamma x}\displaystyle\prod_{n=1}^\infty\left(1+\frac xn\right)e^{-x/n}$. Multiplying the reciprocals for $x$ and $-x$: $$ \frac1{\Gamma(x)\Gamma(-x)} = -x^2e^{\gamma x}e^{-\gamma x}\prod_{n=1}^\infty\left(1+\frac xn\right)\left(1-\frac xn\right)e^{-x/n}e^{x/n} = -x^2\prod_{n=1}^\infty\left(1-\frac{x^2}{n^2}\right). $$ The $e^{\pm\gamma x}$ and $e^{\mp x/n}$ factors cancel exactly — this is precisely the step that a naive symbolic manipulation tends to skip.
Step 3 · Identify with the sine product and conclude
Since $\sin(\pi x)=\pi x\prod_{n=1}^\infty(1-x^2/n^2)$: $$ \frac1{\Gamma(x)\Gamma(-x)} = -x^2\cdot\frac{\sin(\pi x)}{\pi x} = -\frac{x\sin(\pi x)}\pi \implies \Gamma(x)\Gamma(-x) = -\frac\pi{x\sin(\pi x)}. $$ Using the functional equation $\Gamma(1-x)=(-x)\Gamma(-x)$, i.e. $\Gamma(-x)=-\Gamma(1-x)/x$: $$ \Gamma(x)\cdot\left(-\frac{\Gamma(1-x)}x\right) = -\frac\pi{x\sin(\pi x)} \implies \boxed{\Gamma(x)\Gamma(1-x)=\frac\pi{\sin(\pi x)}} $$ (Numerical check at $x=0.3$: $\Gamma(0.3)\Gamma(0.7)\approx3.883222$, matching $\pi/\sin(0.3\pi)\approx3.883222$.)
Summary Table
| Property | Value |
|---|---|
| Cosecant expansion | $1/\sin x=1/x+2x\sum(-1)^k/(x^2-k^2\pi^2)$ |
| Reflection formula | $\Gamma(x)\Gamma(1-x)=\pi/\sin(\pi x)$ |
| Special case $x=1/2$ | $\Gamma(1/2)=\sqrt\pi$ |
Additional Notes
- Correction: a draft derivation of the reflection formula attempted a direct manipulation of the Weierstrass product from Problem 387 but stalled at "after careful analysis... this equals $\sin(\pi x)/\pi$" without carrying out the algebra. The derivation above (Steps 2–3) completes this rigorously by using the canonical form $1/\Gamma(x)=xe^{\gamma x}\prod(1+x/n)e^{-x/n}$, in which the Euler–Mascheroni exponential factors are designed to cancel exactly when $\Gamma(x)$ and $\Gamma(-x)$ are multiplied — verified independently to match both sides numerically at $x=0.3$ and $x=1/2$.
- Setting $x=1/2$ recovers the celebrated special value $\Gamma(1/2)=\sqrt\pi$ directly: $[\Gamma(1/2)]^2=\pi/\sin(\pi/2)=\pi$.
- The reflection formula is the Gamma-function analogue of the sine product itself: just as $\sin(\pi x)$ vanishes at every integer, $1/[\Gamma(x)\Gamma(1-x)]=\sin(\pi x)/\pi$ vanishes there too, consistent with $\Gamma$ having poles (not zeros) at the non-positive integers, as established in Problem 387.