2018 Gamma and Zeta

Calculus Level 3

∫ − ∞ ∞ 1 e x 2018 + 1 d x = ( 2 − 2 ⋅ 2 c / a ) ζ ( 1 a ) Γ ( b a ) \int_{-\infty }^{\infty } \frac{1}{e^{x^{2018}}+1} \, dx=\left(2-2\cdot 2^{c/a}\right) \zeta \left(\frac{1}{a}\right) \Gamma \left(\frac{b}{a}\right)

where a , b , c a,b,c are positive integers. Submit a + b + c a+b+c .


The answer is 6054.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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1 solution

敬全 钟
Dec 24, 2017

∫ − ∞ ∞ 1 e x 2018 + 1 d x = 2 ∫ 0 ∞ 1 e x 2018 + 1 d x (by symmetry) = 2 ∫ 0 ∞ 1 − 1 1 + e − x 2018 d x = 2 ∫ 0 ∞ e − x 2018 − e − 2 x 2018 + e − 3 x 2018 − . . . d x (using the series 1 1 + x = 1 − x + x 2 − x 3 + . . . , see (i)) = 2 Γ ( 2019 2018 ) ∑ k = 1 ∞ ( − 1 ) k ( 1 k 1 2018 ) (see (ii)) = ( 2 − 2 × 2 2017 2018 ) Γ ( 2019 2018 ) ζ ( 1 2018 ) (see (iii)) \begin{aligned} \int^{\infty}_{-\infty}\frac{1}{e^{x^{2018}}+1}\ dx&=&2\int^{\infty}_{0}\frac{1}{e^{x^{2018}}+1}\ dx\quad\color{#3D99F6}{\text{(by symmetry)}}\\ &=&2\int^{\infty}_01-\frac{1}{1+e^{-x^{2018}}}\ dx\\ &=&2\int^{\infty}_0e^{-x^{2018}}-e^{-2x^{2018}}+e^{-3x^{2018}}-...\ dx\quad\color{#3D99F6}{\text{(using the series }}\color{#3D99F6}{\frac{1}{1+x}=1-x+x^2-x^3+...}\color{#3D99F6}{\text{, see (i))}}\\ &=&2\Gamma\left(\frac{2019}{2018}\right)\sum^{\infty}_{k=1}(-1)^k\left(\frac{1}{k^{\frac{1}{2018}}}\right)\quad\color{#3D99F6}{\text{(see (ii))}}\\ &=&\left(2-2\times2^{\frac{2017}{2018}}\right)\Gamma\left(\frac{2019}{2018}\right)\zeta\left(\frac{1}{2018}\right)\quad\color{#3D99F6}{\text{(see (iii))}} \end{aligned}

Thus, a = 2018 , b = 2019 , c = 2018 , a=2018, b=2019, c=2018, giving a + b + c = 6054. a+b+c=6054.

(i) Notice that we can use the series expansion here because 0 < e − k x 2018 ⩽ 1 0<e^{-kx^{2018}}\leqslant1 for each positive integer k ⩾ 1 k\geqslant1 and real number x ⩾ 0. x\geqslant0.

(ii) Let k k be a positive integer. Then,

∫ 0 ∞ e − k x 2018 d x = ∫ 0 ∞ 1 2018 ( 1 k ) 1 2018 u − 2017 2018 e − u d u ( u = k x 2018 ⇒ d u d x = 1 2018 ( 1 k ) 1 2018 u − 2017 2018 ) = ( 1 k ) 1 2018 × 1 2018 Γ ( 1 2018 ) (using the definition of gamma integral, Γ ( s ) : = ∫ 0 ∞ t s − 1 e − t d t . ) = ( 1 k ) 1 2018 Γ ( 2019 2018 ) . \begin{aligned} \int^{\infty}_0e^{-kx^{2018}}\ dx&=&\int^{\infty}_0\frac{1}{2018}\left(\frac{1}{k}\right)^{\frac{1}{2018}}u^{-\frac{2017}{2018}}e^{-u}\ du\ \ \ \color{#3D99F6}\left({u=kx^{2018}\Rightarrow\frac{du}{dx}=\frac{1}{2018}\left(\frac{1}{k}\right)^{\frac{1}{2018}}u^{-\frac{2017}{2018}}}\right)\\ &=&\left(\frac{1}{k}\right)^{\frac{1}{2018}}\times\frac{1}{2018}\Gamma\left(\frac{1}{2018}\right)\ \ \ \color{#3D99F6}{\text{(using the definition of gamma integral, }\Gamma(s):=\int^{\infty}_0t^{s-1}e^{-t}\ dt.)}\\ &=&\left(\frac{1}{k}\right)^{\frac{1}{2018}}\Gamma\left(\frac{2019}{2018}\right). \end{aligned} The proof is complete.

(iii) ∑ k = 1 ∞ ( − 1 ) k ( 1 k 1 2018 ) = 1 − 1 2 1 2018 + 1 3 1 2018 − 1 4 1 2018 + . . . = 1 + ( 1 2 1 2018 − 2 2 1 2018 ) + 1 3 1 2018 − ( 1 4 1 2018 − 2 4 1 2018 ) + . . . = ( 1 + 1 2 1 2018 + 1 3 1 2018 + 1 4 1 2018 + . . . ) − 2 ( 1 2 1 2018 + 1 4 1 2018 + 1 6 1 2018 + 1 8 1 2018 . . . ) = ζ ( 1 2018 ) − 1 2 1 2018 × 2 ( 1 + 1 2 1 2018 + 1 3 1 2018 + 1 4 1 2018 . . . ) = ζ ( 1 2018 ) − 2 2 1 2018 ζ ( 1 2018 ) = ( 1 − 1 2 2017 2018 ) ζ ( 1 2018 ) . \begin{aligned} \sum^{\infty}_{k=1}(-1)^k\left(\frac{1}{k^{\frac{1}{2018}}}\right)&=&1{\color{#3D99F6}{-\frac{1}{2^{\frac{1}{2018}}}}}+\frac{1}{3^{\frac{1}{2018}}}{\color{#3D99F6}{-\frac{1}{4^{\frac{1}{2018}}}}}+...\\ &=&1+{\color{#3D99F6}{\left(\frac{1}{2^{\frac{1}{2018}}}-\frac{2}{2^{\frac{1}{2018}}}\right)}}+\frac{1}{3^{\frac{1}{2018}}}-{\color{#3D99F6}{\left(\frac{1}{4^{\frac{1}{2018}}}-\frac{2}{4^{\frac{1}{2018}}}\right)}}+...\\ &=&\left(1+\frac{1}{2^{\frac{1}{2018}}}+\frac{1}{3^{\frac{1}{2018}}}+\frac{1}{4^{\frac{1}{2018}}}+...\right)-\color{magenta}{2\left(\frac{1}{2^{\frac{1}{2018}}}+\frac{1}{4^{\frac{1}{2018}}}+\frac{1}{6^{\frac{1}{2018}}}+\frac{1}{8^{\frac{1}{2018}}}...\right)}\\ &=&\zeta\left(\frac{1}{2018}\right)-\frac{1}{2^{\frac{1}{2018}}}\times\color{magenta}{2\left(1+\frac{1}{2^{\frac{1}{2018}}}+\frac{1}{3^{\frac{1}{2018}}}+\frac{1}{4^{\frac{1}{2018}}}...\right)}\\ &=&\zeta\left(\frac{1}{2018}\right)-\frac{2}{2^{\frac{1}{2018}}}\zeta\left(\frac{1}{2018}\right)\\ &=&\left(1-\frac{1}{2^{\frac{2017}{2018}}}\right)\zeta\left(\frac{1}{2018}\right). \end{aligned}

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