Forgot password? New user? Sign up
Existing user? Log in
∫01(lnΓ(x))sin(πx) dx=1π(1+lnπA)\large \int_0^1(\ln\Gamma(x))\sin(\pi x)\ dx=\frac1\pi\left(1+\ln{\frac\pi A}\right)∫01(lnΓ(x))sin(πx) dx=π1(1+lnAπ)
Find AAA.
Notation: Γ(⋅)\Gamma(\cdot)Γ(⋅) denotes the gamma function.
Try a similar problem here.
Problem Loading...
Note Loading...
Set Loading...