Like the question states, how do I evaluate the integral
∫∞−∞1ex2+1dx
I know of no methods to evaluate this, but when I plug this into Mathematica and Wolfram Alpha, it returns
∫∞−∞1ex2+1dx=(1−√2)√πζ(12)
where ζ(x) is the Riemann Zeta Function.
Answer
The given integral equals
2∫+∞0dx1+ex2=∫+∞0dz√z(1+ez)=∑n≥1(−1)n+1∫+∞0e−nz√zdz
or
√π∑n≥1(−1)n+1√n=√πη(12)=(1−√2)√πζ(12)
as claimed by WA.
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