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Question Number 62415 by mathmax by abdo last updated on 20/Jun/19
calculate f(x,y) =∫_0 ^∞ e^(−xt) ln(yt) dt with x>0 and y>0 .
calculatef(x,y)=∫0∞e−xtln(yt)dtwithx>0andy>0.
Commented by mathmax 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.
wehavef(x,y)=∫0∞e−xt(lny+ln(t)dt=ln(y)∫0∞e−xtdt+∫0∞e−xtln(t)dt∫0∞e−xtdt=[−1xe−xt]t=0∞=1x∫0∞e−xtln(t)dt=xt=u∫0∞e−uln(ux)dux=1x{∫0∞e−uln(u)du−ln(x)∫0∞e−udu}=1x{−γ−ln(x)×1}=−1x{ln(x)+γ)⇒f(x,y)=ln(y)x−1x{ln(x)+γ}=1x(ln(yx)−γ)γiseulerconstantnumber.

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