calculate f(x,y) =∫_0 ^∞ e^(−xt) ln(yt) dt with x>0 and y>0 .


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Question Number 62415 by mathmax by abdo last updated on 20/Jun/19

$c a l c u l a t e f (x, y) = \int_{0}^{\infty} e^{- x t} l n (y t) d t w i t h x > 0 a n d y > 0 .$
Commented bymathmax by abdo last updated on 23/Jun/19
$we have f(x,y) =∫_0 ^∞ e^(−xt) (lny +ln(t)dt =ln(y)∫_0 ^∞ e^(−xt) dt +∫_0 ^∞ e^(−xt) ln(t)dt ∫_0 ^∞ e^(−xt) dt =[−(1/x)e^(−xt) ]_(t=0) ^∞ =(1/x) ∫_0 ^∞ e^(−xt) ln(t) dt =_(xt =u) ∫_0 ^∞ e^(−u) ln((u/x))(du/x) =(1/x){ ∫_0 ^∞ e^(−u) ln(u)du−ln(x)∫_0 ^∞ e^(−u) du} =(1/x){ −γ −ln(x)×1} =−(1/x){ln(x)+γ) ⇒ f(x,y) =((ln(y))/x) −(1/x){ln(x)+γ} =(1/x)( ln((y/x))−γ) γ is euler constant number.$
$w e h a v e f (x, y) = \int_{0}^{\infty} e^{- x t} (l n y + l n (t) d t = l n (y) \int_{0}^{\infty} e^{- x t} d t + \int_{0}^{\infty} e^{- x t} l n (t) d t$ $\int_{0}^{\infty} e^{- x t} d t = {[- \frac{1}{x} e^{- x t}]}_{t = 0}^{\infty} = \frac{1}{x}$ $\int_{0}^{\infty} e^{- x t} l n (t) d t =_{x t = u} \int_{0}^{\infty} e^{- u} l n (\frac{u}{x}) \frac{d u}{x}$ $= \frac{1}{x} {\int_{0}^{\infty} e^{- u} l n (u) d u - l n (x) \int_{0}^{\infty} e^{- u} d u}$ $= \frac{1}{x} {- γ - l n (x) \times 1} = - \frac{1}{x} {l n (x) + γ) \Rightarrow$ $f (x, y) = \frac{l n (y)}{x} - \frac{1}{x} {l n (x) + γ} = \frac{1}{x} (l n (\frac{y}{x}) - γ)$ $γ i s e u l e r c o n s t a n t n u m b e r .$
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