calculate determinant ((( L ( ∫_1 ^( ∞) (( e^( −tx) )/x)dx ) =


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Question Number 217219 by mnjuly1970 last updated on 06/Mar/25

$c a l c u l a t e$ $\begin{matrix} L (\int_{1}^{\infty} \frac{e^{- t x}}{x} d x) \underset{transfom}{\overset{laplace}{=}} ?; t > 0 \end{matrix}$
	Answered by Wuji last updated on 06/Mar/25
	$f(t)=∫_1 ^∞ (e^(−tx) /x)dx , t>0 u=tx ⇒dx=(1/t)du ; x=(u/t) f(t)=∫_t ^∞ (e^(−u) /(u/t))•(1/t)du =∫_t ^∞ (e^(−u) /u)du=E_1 (t) f(t)=E_1 (t) L{f}(s) =∫_0 ^∞ e^(−st) E_1 (t) L{f}(s)=∫_0 ^∞ e^(−st) [∫_0 ^∞ (e^(−tx) /x)dx]dt =∫_1 ^∞ (1/x)[∫_0 ^∞ e^(−t(s+x)) dt]dx ∫_0 ^∞ e^(−t(s+x)) dt=(1/(s+x)) {∀R(s+x) >0} L{f}(s)=∫_1 ^∞ (1/(x(s+x)))dx (1/(x(s+x)))=(1/s)((1/x)−(1/(s+x))) L{f}(s)=(1/s)∫_1 ^∞ (1/x)dx−∫_1 ^∞ (1/(s+x))dx ∫_1 ^R (1/x)dx =ln(R) ; ∫^R _1 (2/(s+x))dx=∫_(s+1) ^(s+R) (1/u)du=ln(s+R)−ln(s+1) ∫_1 ^R (1/x)dx−∫_1 ^R (1/(s+x))dx=ln(R)−[ln(s+R)−ln(s+1)]=ln(((R(s+1))/((s+R)))) ((R(s+1))/((s+R))) → s+1 ln(s+1) ∴L{f}(s)=(1/s)ln(s+1) f(t)=∫_0 ^∞ (e^(−tx) /x)dt=(1/s)ln(s+1)$
	$f (t) = {\int^{\infty}}_{1} \frac{e^{- tx}}{x} dx, t > 0$ $u = tx \Rightarrow dx = \frac{1}{t} du; x = \frac{u}{t}$ $f (t) = \int_{t}^{\infty} \frac{e^{- u}}{\frac{u}{t}} ∙ \frac{1}{t} du = \int_{t}^{\infty} \frac{e^{- u}}{u} du = E_{1} (t)$ $f (t) = E_{1} (t)$ $L {f} (s) = \int_{0}^{\infty} e^{- st} E_{1} (t)$ $L {f} (s) = \underset{0}{\int^{\infty}} e^{- st} [\underset{0}{\int^{\infty}} \frac{e^{- tx}}{x} dx] dt$ $= \underset{1}{\int^{\infty}} \frac{1}{x} [{\int^{\infty}}_{0} e^{- t (s + x)} dt] dx$ $\underset{0}{\int^{\infty}} e^{- t (s + x)} dt = \frac{1}{s + x} {\forall R (s + x) > 0}$ $L {f} (s) = \int_{1}^{\infty} \frac{1}{x (s + x)} dx$ $\frac{1}{x (s + x)} = \frac{1}{s} (\frac{1}{x} - \frac{1}{s + x})$ $L {f} (s) = \frac{1}{s} \underset{1}{\int^{\infty}} \frac{1}{x} dx - \underset{1}{\int^{\infty}} \frac{1}{s + x} dx$ $\underset{1}{\int^{R}} \frac{1}{x} dx = \ln (R); {\int_{1}}^{R} \frac{2}{s + x} dx = \underset{s + 1}{\int^{s + R}} \frac{1}{u} du = \ln (s + R) - \ln (s + 1)$ $\underset{1}{\int^{R}} \frac{1}{x} dx - \underset{1}{\int^{R}} \frac{1}{s + x} dx = \ln (R) - [\ln (s + R) - \ln (s + 1)] = \ln (\frac{R (s + 1)}{(s + R)})$ $\frac{R (s + 1)}{(s + R)} \to s + 1$ $\ln (s + 1)$ $∴ L {f} (s) = \frac{1}{s} \ln (s + 1)$ $f (t) = \underset{0}{\int^{\infty}} \frac{e^{- tx}}{x} dt = \frac{1}{s} \ln (s + 1)$
	Commented by mnjuly1970 last updated on 06/Mar/25

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