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Question Number 130431 by mnjuly1970 last updated on 25/Jan/21

$. . . n i c e c a l c u l u s . . .$ $p l e a s e e v a l u a t e ::$ $ϕ = \int_{0}^{\infty} t a n h (x) . e^{- s x} d x = ? ?$ $(s > 0 a n d r e a l . . .)$
	Answered by Dwaipayan Shikari last updated on 25/Jan/21

	$\int_{0}^{\infty} \frac{e^{2 x} - 1}{e^{2 x} + 1} e^{- s x} d x$ $= \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \int_{0}^{\infty} e^{- 2 n x} e^{2 x - s x} - e^{- 2 n x - s x} d x$ $= \underset{n = 1}{\sum^{\infty}} {(- 1)}^{n + 1} \frac{1}{(2 n + s - 2)} - \frac{1}{(2 n + s)}$ $= (\frac{1}{s} - \frac{1}{2 + s} + \frac{1}{4 + s} - \frac{1}{6 + s} - . .) - (\frac{1}{2 + s} - \frac{1}{4 + s} + . . .)$ $= 2 (\frac{1}{s} + \frac{1}{4 + s} + \frac{1}{8 + s} + . . .) - 2 (\frac{1}{2 + s} + \frac{1}{6 + s} + \frac{1}{10 + s} + . . .) - \frac{1}{s}$ $= \frac{1}{2} (\underset{n = 0}{\sum^{\infty}} \frac{1}{n + \frac{s}{4}} - \underset{n = 0}{\sum^{\infty}} \frac{1}{n + \frac{1}{2} + \frac{s}{4}}) - \frac{1}{s}$ $= \frac{1}{2} (ψ (\frac{1}{2} + \frac{s}{4}) - ψ (\frac{s}{4})) - \frac{1}{s}$
	Commented bymnjuly1970 last updated on 25/Jan/21

	$p e r f e c t m r p a y a n . .$
	Commented byDwaipayan Shikari last updated on 25/Jan/21

	$f o r s = 1, \frac{1}{2} (ψ (\frac{3}{4}) - ψ (\frac{1}{4})) - \frac{1}{s} = \frac{1}{2} \int_{0}^{1} \frac{x^{- \frac{3}{4}} - x^{- \frac{1}{4}}}{1 - x} d x - \frac{1}{s}, x = u^{4}$ $= 2 \int_{0}^{1} \frac{1 - u^{2}}{1 - u^{4}} - \frac{1}{1} d u = 2 . \frac{π}{4} - 1 = \frac{π}{2} - 1$
	Commented bymnjuly1970 last updated on 25/Jan/21

	$t h a n k s a l o t . . .$
	Answered by mnjuly1970 last updated on 26/Jan/21
	$φ=∫_0 ^( ∞) ((e^(2x) −1)/(e^(2x) +1))∗e^(−sx) dx =_(x:=−ln((√t) )) ^(2x:=−ln(t)) ∫_0 ^( 1) (((1/t)−1)/((1/t)+1))∗t^(s/2) ((1/(2t)))dt =(1/2)∫_0 ^( 1) ((1−t)/(1+t))∗t^((s/2)−1) dt =(1/2)∫_0 ^( 1) (((1−2t+t^2 ))/(1−t^2 ))∗t^((s/2)−1) dt =(1/2)∫_0 ^( 1) ((t^((s/2)−1) −2t^(s/2) +t^((s/2)+1) )/(1−t^2 ))dt =^(t^2 =y) (1/2)∫_0 ^( 1) ((y^((s/4)−(1/2)) −2y^(s/4) +y^((s/4)+(1/2)) )/(1−y))((1/2))y^((−1)/2) dy =(1/4)∫_0 ^( 1) ((y^((s/4)−1) −1+1−y^((s/4)−(1/2)) +y^(s/4) −1+1−y^((s/4)−(1/2)) )/(1−y))dy =(1/4){(γ−∫_0 ^( 1) ((1−y^((s/4)−1) )/(1−y))dy )+(−γ+∫((1−y^((s/4)−(1/2)) )/(1−y))dy)+(γ−∫_0 ^( 1) ((1−y^(s/4) )/(1−y))dy)+(−γ+∫_0 ^( 1) ((1−y^((s/4)−(1/2)) )/(1−y))dy)} =(1/4)(−ψ((s/4))+2ψ((s/4)+(1/2))−ψ((s/4)+1)) =(1/4)(−ψ((s/4))+2ψ((s/4)+(1/2))−ψ((s/4))−(4/s)) =(1/2)(ψ((s/4)+(1/2))−ψ((s/4)))−(1/s) ∴ example: φ(1)=(1/2)(ψ((3/4))−ψ((1/4)))−1 we know: ψ(s)−ψ(1−s)=−πcot(πs) ψ((3/4))−ψ((1/4))=(−π)(−1)=π φ(1)=(π/2) −1 ....✓$
	$ϕ = \int_{0}^{\infty} \frac{e^{2 x} - 1}{e^{2 x} + 1} * e^{- s x} d x$ $\underset{x := - l n (\sqrt{t})}{\overset{2 x := - l n (t)}{=}} \int_{0}^{1} \frac{\frac{1}{t} - 1}{\frac{1}{t} + 1} * t^{\frac{s}{2}} (\frac{1}{2 t}) d t$ $= \frac{1}{2} \int_{0}^{1} \frac{1 - t}{1 + t} * t^{\frac{s}{2} - 1} d t$ $= \frac{1}{2} \int_{0}^{1} \frac{(1 - 2 t + t^{2})}{1 - t^{2}} * t^{\frac{s}{2} - 1} d t$ $= \frac{1}{2} \int_{0}^{1} \frac{t^{\frac{s}{2} - 1} - 2 t^{\frac{s}{2}} + t^{\frac{s}{2} + 1}}{1 - t^{2}} d t$ $\overset{t^{2} = y}{=} \frac{1}{2} \int_{0}^{1} \frac{y^{\frac{s}{4} - \frac{1}{2}} - 2 y^{\frac{s}{4}} + y^{\frac{s}{4} + \frac{1}{2}}}{1 - y} (\frac{1}{2}) y^{\frac{- 1}{2}} d y$ $= \frac{1}{4} \int_{0}^{1} \frac{y^{\frac{s}{4} - 1} - 1 + 1 - y^{\frac{s}{4} - \frac{1}{2}} + y^{\frac{s}{4}} - 1 + 1 - y^{\frac{s}{4} - \frac{1}{2}}}{1 - y} d y$ $= \frac{1}{4} {(γ - \int_{0}^{1} \frac{1 - y^{\frac{s}{4} - 1}}{1 - y} d y) + (- γ + \int \frac{1 - y^{\frac{s}{4} - \frac{1}{2}}}{1 - y} d y) + (γ - \int_{0}^{1} \frac{1 - y^{\frac{s}{4}}}{1 - y} d y) + (- γ + \int_{0}^{1} \frac{1 - y^{\frac{s}{4} - \frac{1}{2}}}{1 - y} d y)}$ $= \frac{1}{4} (- ψ (\frac{s}{4}) + 2 ψ (\frac{s}{4} + \frac{1}{2}) - ψ (\frac{s}{4} + 1))$ $= \frac{1}{4} (- ψ (\frac{s}{4}) + 2 ψ (\frac{s}{4} + \frac{1}{2}) - ψ (\frac{s}{4}) - \frac{4}{s})$ $= \frac{1}{2} (ψ (\frac{s}{4} + \frac{1}{2}) - ψ (\frac{s}{4})) - \frac{1}{s}$ $∴ e x a m p l e : ϕ (1) = \frac{1}{2} (ψ (\frac{3}{4}) - ψ (\frac{1}{4})) - 1$ $w e k n o w :$ $ψ (s) - ψ (1 - s) = - π c o t (π s)$ $ψ (\frac{3}{4}) - ψ (\frac{1}{4}) = (- π) (- 1) = π$ $ϕ (1) = \frac{π}{2} - 1 . . . . ✓$
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