Harmonic integral

Calculus Level 4

1 / 2 0 H x d x = a b γ c d ln π \large \int_{-1/2}^0 H_ x \, dx = \dfrac ab \gamma - \dfrac cd \ln \pi

If the equation above holds true for positive integers a , b , c a,b,c and d d with gcd ( a , b ) = gcd ( c , d ) = 1 \gcd(a,b) = \gcd(c,d) = 1 , find a + b + c + d a+b+c+d .

Notations :


The answer is 6.

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1 solution

Aareyan Manzoor
Mar 19, 2016

We use the digamma function

H x d x = ψ ( x + 1 ) + γ d x = ln ( Γ ( x + 1 ) ) + γ x + C \int H_x dx=\int \psi(x+1)+\gamma dx=\ln(\Gamma(x+1)) +\gamma x+C so, 1 / 2 0 H x d x = ln ( Γ ( 1 ) ) ( ln ( Γ ( 1 2 ) ) 1 2 γ ) = 1 2 γ 1 2 ln π \int_{-1/2}^0 H_x dx=\ln(\Gamma(1))-\left(\ln\left(\Gamma\left(\dfrac{1}{2}\right)\right)-\dfrac{1}{2}\gamma\right)\\=\dfrac{1}{2}\gamma-\dfrac{1}{2}\ln \pi

Same way! +1

Nihar Mahajan - 5 years, 2 months ago

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