# Question #173f9

Nov 26, 2017

$A = L n 3$ and ${e}^{A} = 3$

#### Explanation:

$A = {\int}_{1}^{\infty} \left(\frac{1}{u} ^ 2 - \frac{1}{u} ^ 4\right) \cdot \frac{\mathrm{du}}{L} \nu$

After using $x = L \nu$, $u = {e}^{x}$ and $\mathrm{du} = {e}^{x} \cdot \mathrm{dx}$ transforms, $A$ became,

$A = {\int}_{0}^{\infty} \left[\frac{1}{{e}^{x}} ^ 2 - \frac{1}{{e}^{x}} ^ 4\right] \cdot \frac{{e}^{x} \cdot \mathrm{dx}}{x}$

=${\int}_{0}^{\infty} \left({e}^{- 2 x} - {e}^{- 4 x}\right) \cdot \frac{{e}^{x} \cdot \mathrm{dx}}{x}$

=${\int}_{0}^{\infty} \frac{\left({e}^{- x} - {e}^{- 3 x}\right) \cdot \mathrm{dx}}{x}$

Now, I solved this integral via Feynman's trick by choosing $b = 3$

$A \left(b\right) = {\int}_{0}^{\infty} \frac{\left({e}^{- x} - {e}^{- b x}\right) \cdot \mathrm{dx}}{x}$

After differentiation both sides according to $b$

$A ' \left(b\right) = {\int}_{0}^{\infty} {e}^{- b x} \cdot \mathrm{dx}$

=${\left[- \frac{1}{b} \cdot {e}^{- b x}\right]}_{0}^{\infty}$

=$\frac{1}{b}$

After integrating both sides,

$A \left(b\right) = L n b + C$

Now I chose a strategic value $b = {b}_{0}$ in order to make our integrand vanish so that $A \left({b}_{0}\right) = 0$. In this case, take $b = 1$ so that $A \left(1\right) = 0$. Consequently,

$L n 1 + C = 0$ or, $C = 0$

Hence, I found value of $A$ after setting $b = 3$

$A = A \left(3\right) = \ln 3$

Thus value of ${e}^{A}$ is ${e}^{L n 3} = 3$