Explicit f(x)= ∫_1 ^∞ ((lnt)/((x^2 +t^2 )^2 )) dt


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Question Number 73293 by ~blr237~ last updated on 10/Nov/19

$E x p l i c i t f (x) = \int_{1}^{\infty} \frac{l n t}{{(x^{2} + t^{2})}^{2}} d t$
Commented by mathmax by abdo last updated on 10/Nov/19
$let w(x)=∫_1 ^∞ ((lnt)/(x^2 +t^2 ))dt ⇒w(x)=_(t=(1/u)) −∫_0 ^1 ((−lnu)/(x^2 +(1/u^2 )))(−(du/u^2 )) =−∫_0 ^1 ((lnu)/(u^2 x^2 +1))du =_(ux=z) −∫_0 ^x ((ln((z/x)))/(z^2 +1))(dz/x) =−(1/x)∫_0 ^x ((lnz−lnx)/(z^2 +1))dz =−(1/x)∫_0 ^x ((lnz)/(z^2 +1))dz+((lnx)/x) arctanx w(x)=((lnx)/x)arctanx −(1/x)∫_0 ^x ((lnz)/(z^2 +1))dz and w^′ (x)= −∫_1 ^(+∞) ((2xlnt)/((x^2 +t^2 )^2 ))dt =−2xf(x) ⇒f(x)=−(1/(2x))w^′ (x) we have w^′ (x)=((1−lnx)/x^2 ) arctanx +((lnx)/(x(1+x^2 ))) +(1/x^2 ) ∫_0 ^x ((lnz)/(z^2 +1))dz −((lnx)/(x(x^2 +1))) ⇒ f(x)=−(1/(2x)){((1−lnx)/x^2 ) arctanx +(1/x^2 ) ∫_0 ^x ((lnz)/(z^2 +1))dz} =((lnx−1)/(2x^3 )) arctanx −(1/(2x^3 )) ∫_0 ^x ((lnz)/(z^2 +1))dz$
$l e t w (x) = \int_{1}^{\infty} \frac{l n t}{x^{2} + t^{2}} d t \Rightarrow w (x) =_{t = \frac{1}{u}} - \int_{0}^{1} \frac{- l n u}{x^{2} + \frac{1}{u^{2}}} (- \frac{d u}{u^{2}})$ $= - \int_{0}^{1} \frac{l n u}{u^{2} x^{2} + 1} d u =_{u x = z} - \int_{0}^{x} \frac{l n (\frac{z}{x})}{z^{2} + 1} \frac{d z}{x}$ $= - \frac{1}{x} \int_{0}^{x} \frac{l n z - l n x}{z^{2} + 1} d z = - \frac{1}{x} \int_{0}^{x} \frac{l n z}{z^{2} + 1} d z + \frac{l n x}{x} a r c t a n x$ $w (x) = \frac{l n x}{x} a r c t a n x - \frac{1}{x} \int_{0}^{x} \frac{l n z}{z^{2} + 1} d z a n d$ $w^{^{'}} (x) = - \int_{1}^{+ \infty} \frac{2 x l n t}{{(x^{2} + t^{2})}^{2}} d t = - 2 x f (x) \Rightarrow f (x) = - \frac{1}{2 x} w^{^{'}} (x) w e h a v e$ $w^{^{'}} (x) = \frac{1 - l n x}{x^{2}} a r c t a n x + \frac{l n x}{x (1 + x^{2})} + \frac{1}{x^{2}} \int_{0}^{x} \frac{l n z}{z^{2} + 1} d z - \frac{l n x}{x (x^{2} + 1)} \Rightarrow$ $f (x) = - \frac{1}{2 x} {\frac{1 - l n x}{x^{2}} a r c t a n x + \frac{1}{x^{2}} \int_{0}^{x} \frac{l n z}{z^{2} + 1} d z}$ $= \frac{l n x - 1}{2 x^{3}} a r c t a n x - \frac{1}{2 x^{3}} \int_{0}^{x} \frac{l n z}{z^{2} + 1} d z$
Commented by ~blr237~ last updated on 10/Nov/19

$t h e f i r s t o n e$
	Answered by mind is power last updated on 10/Nov/19

	$check it sir$ $f (x) = \int_{1}^{+ \infty} \frac{\ln (t)}{{(x^{2} + t^{2})}^{2}} dt or \int_{0}^{+ \infty} \frac{\ln (t)}{{(x^{2} + t^{2})}^{2}} ?$
	Commented by ~blr237~ last updated on 10/Nov/19

	$t h e f i r s t o n e$
	Commented by ~blr237~ last updated on 10/Nov/19

	$t h e f i r s t o n e$
	Answered by mind is power last updated on 10/Nov/19
	$f(x)=∫_1 ^(+∞) ((ln(t))/((x^2 +t^2 )^2 ))dt f(x)=(d/da)∫((−ln(t))/((a+t^2 )))dt∣_(a=x^2 ) a>0 let g(a)=−∫_1 ^(+∞) ((ln(t))/(a+t^2 ))dt t=(1/u)⇒dt=−(du/u^2 )⇒g(a)=∫_0 ^1 ((ln(u))/(au^2 +1))du =∫_0 ^1 ((ln(u))/((u(√a)−i)(u(√a)+i)))du =∫_0 ^1 ((ln(u)du)/(2i(u(√a)−i)))−(1/(2i))∫_0 ^1 ((ln(u)du)/((u(√a)+i))) =∫_0 ^1 ((ln(u)du)/(−2(−iu(√a)−1)))+(1/2)∫_0 ^1 ((ln(u)du)/((iu(√a)−1))) −iu(√a)=x in first iu(√a)=y in 2nd ⇒∫_0 ^(−i(√a)) ((ln((x/(√a))i))/(−2(x−1))).(dx/(−i(√a)))+(1/2)∫_0 ^(i(√a)) ((ln((y/(i(√a)))))/((y−1))).(dy/(i(√a))) =−(1/(2i(√a))){∫_0 ^(−i(√a)) ((ln(x)+ln((√a))+((iπ)/2))/(1−x))dx+∫_0 ^(i(√a)) ((ln(y)−ln((√a))−((iπ)/2))/(1−y))dy} use ∫_1 ^(1−z) ((ln(t))/(1−x))dt=Li_2 (z) =−(1/(2i(√a))){∫_(1−1) ^(1−(1+i(√(a)))) ((ln(x))/(1−x))dx+∫_(1−1) ^(1−(1−i(√a))) ((ln(y))/(1−y))dy−(ln(√a)+i(π/2))log(1+i(√a))+(ln(√a)+((iπ)/2))log(1−i(√(a)})) =−(1/(2i(√a)))Li_2 (1+i(√a))−((Li_2 (1−i(√a)))/(2i(√a)))+((ln((√a))+i(π/2))/(2i(√a)))log(((1+i(√a))/(1−i(√a))))+c f(x)=(d/da){−(1/(2i(√a)))Li_2 (1+i(√a))−((Li_2 (1−i(√a)))/(2i(√a)))+((ln((√a))+i(π/2))/(2i(√a)))log(((1+i(√a))/(1−i(√a))))+c}_(a=x^2 ) withe (d/dz)Li_2 (z)=−((ln(1−z))/z) you will get close forme littel big! we can use ∀z∈C −{i}(1/(2i)).log(((1+iz)/(1−iz)))=arctan(z) f(x)=(d/da){−((Li_2 (1+i(√a)))/(2i(√a)))−((Li_2 (1−i(√a)))/(2i(√a)))+((ln((√a))+((iπ)/2))/(√a)).arctan((√a))}∣_(a=x^2 )$
	$f (x) = \int_{1}^{+ \infty} \frac{\ln (t)}{{(x^{2} + t^{2})}^{2}} dt$ $f (x) = \frac{d}{da} \int \frac{- \ln (t)}{(a + t^{2})} dt ∣_{a = x^{2}}$ $a > 0$ $let g (a) = - \int_{1}^{+ \infty} \frac{\ln (t)}{a + t^{2}} dt$ $t = \frac{1}{u} \Rightarrow dt = - \frac{du}{u^{2}} \Rightarrow g (a) = \int_{0}^{1} \frac{\ln (u)}{{au}^{2} + 1} du$ $= \int_{0}^{1} \frac{\ln (u)}{(u \sqrt{a} - i) (u \sqrt{a} + i)} du$ $= \int_{0}^{1} \frac{\ln (u) du}{2 i (u \sqrt{a} - i)} - \frac{1}{2 i} \int_{0}^{1} \frac{\ln (u) du}{(u \sqrt{a} + i)}$ $= \int_{0}^{1} \frac{\ln (u) du}{- 2 (- iu \sqrt{a} - 1)} + \frac{1}{2} \int_{0}^{1} \frac{\ln (u) du}{(iu \sqrt{a} - 1)}$ $- iu \sqrt{a} = x in first iu \sqrt{a} = y in 2 nd$ $\Rightarrow \int_{0}^{- i \sqrt{a}} \frac{\ln (\frac{x}{\sqrt{a}} i)}{- 2 (x - 1)} . \frac{dx}{- i \sqrt{a}} + \frac{1}{2} \int_{0}^{i \sqrt{a}} \frac{\ln (\frac{y}{i \sqrt{a}})}{(y - 1)} . \frac{dy}{i \sqrt{a}}$ $= - \frac{1}{2 i \sqrt{a}} {\int_{0}^{- i \sqrt{a}} \frac{\ln (x) + \ln (\sqrt{a}) + \frac{i π}{2}}{1 - x} dx + \int_{0}^{i \sqrt{a}} \frac{\ln (y) - \ln (\sqrt{a}) - \frac{i π}{2}}{1 - y} dy}$ $use \int_{1}^{1 - z} \frac{\ln (t)}{1 - x} dt = {Li}_{2} (z)$ $= - \frac{1}{2 i \sqrt{a}} {\int_{1 - 1}^{1 - (1 + i \sqrt{a)}} \frac{\ln (x)}{1 - x} dx + \int_{1 - 1}^{1 - (1 - i \sqrt{a})} \frac{\ln (y)}{1 - y} dy - (\ln \sqrt{a} + i \frac{π}{2}) \log (1 + i \sqrt{a}) + (\ln \sqrt{a} + \frac{i π}{2}) \log (1 - i \sqrt{a)}}$ $= - \frac{1}{2 i \sqrt{a}} {Li}_{2} (1 + i \sqrt{a}) - \frac{{Li}_{2} (1 - i \sqrt{a})}{2 i \sqrt{a}} + \frac{\ln (\sqrt{a}) + i \frac{π}{2}}{2 i \sqrt{a}} \log (\frac{1 + i \sqrt{a}}{1 - i \sqrt{a}}) + c$ $f (x) = \frac{d}{da} {- \frac{1}{2 i \sqrt{a}} {Li}_{2} (1 + i \sqrt{a}) - \frac{{Li}_{2} (1 - i \sqrt{a})}{2 i \sqrt{a}} + \frac{\ln (\sqrt{a}) + i \frac{π}{2}}{2 i \sqrt{a}} \log (\frac{1 + i \sqrt{a}}{1 - i \sqrt{a}}) + c}_{a = x^{2}}$ $withe \frac{d}{dz} {Li}_{2} (z) = - \frac{\ln (1 - z)}{z}$ $you will get close forme littel big!$ $we can use \forall z \in C - {i} \frac{1}{2 i} . \log (\frac{1 + iz}{1 - iz}) = \arctan (z)$ $f (x) = \frac{d}{da} {- \frac{{Li}_{2} (1 + i \sqrt{a})}{2 i \sqrt{a}} - \frac{{Li}_{2} (1 - i \sqrt{a})}{2 i \sqrt{a}} + \frac{\ln (\sqrt{a}) + \frac{i π}{2}}{\sqrt{a}} . \arctan (\sqrt{a})} ∣_{a = x^{2}}$
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