"I am Back" says Integration Part 6

Calculus Level 5

xe2xee2xdx\large \int_{-\infty}^{\infty} {xe^{2x}e^{-{e}^{2x}} \, dx}

The above Integral can be expressed as γa,-\dfrac{\gamma}{a},

where γ\gamma denotes the Euler-Mascheroni constant γ=limn(lnn+k=1n1k)0.5772.\displaystyle \gamma = \lim_{n\to\infty} \left( - \ln n + \sum_{k=1}^n \dfrac1k \right) \approx 0.5772 .

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