Triple Integral

$Using$ $partial$ $fraction$ $for$ ${\bf \frac{1}{(X - J - M){(X - J - M - 1})} }$ $we$ $obtain:$

${\bf \frac{1}{(X - J - M){(X - J - M - 1})} = \frac{A}{X - J - M} + \frac{B}{X - J - M - 1} \implies }$

${\bf 1 = (A + B) * X - (J + M + 1) * A - (J + M) * B \implies }$

${\bf A + B = 0 }$

${\bf (J + M + 1) * A + (J + M) * B = -1 }$ ${\bf \implies }$

${\bf A = -1 }$ $and$ ${\bf B = 1 \implies }$

${\bf \sum_{J = 1}^{N} \frac{1}{(X - J - M){(X - J - M - 1})} = }$

${\bf \sum_{J = 1}^{N} (\frac{1}{X - J - M - 1} - \frac{1}{X - J - M}) = }$

${\bf (\frac{1}{X - M - 2} - \frac{1}{X - M - 1}) + (\frac{1}{X - M - 3} - \frac{1}{X - M - 2}) }$ + ... +

${\bf (\frac{1}{X - M + 1 - N} - \frac{1}{X - M + 2 - N}) + (\frac{1}{X - M - N} - \frac{1}{X - M + 1 - N}) + (\frac{1}{X - M - 1 - N} - \frac{1}{X - M - N}) = }$

${\bf \frac{1}{X - M - 1 - N} - \frac{1}{X - M - 1} \implies }$

${\bf \int_{X = 2N + M + 1}^{\infty} \frac{1}{(X - J - M){(X - J - M - 1})} dx = }$ ${\bf \int_{X = 2N + M + 1}^{\infty} (\frac{1}{X - M - 1 - N} - \frac{1}{X - M - 1}) = }$

${\bf \ln(1 - \frac{N}{X - M - 1})|_{X = 2N + M + 1}^{\infty} = ln(2) \implies }$

${\bf ln(2) * \int_{Y = e}^{Y = e^2} \frac{1/Y}{ln(Y)} dy = }$

${\bf ln(2) * ln(ln(y))|_{e}^{e^2} = (ln(2))^2 \implies }$

${\bf (ln(2))^2 * \int_{Z = e^e}^{Z = e^{e^2}} \frac{(1/(Z * ln(Z))}{ln(ln(Z))} dz = }$ ${\bf (ln(2))^2 * (ln(ln(ln(Z))|_{e^e}^{e^{e^2}} = (ln(2))^3 }$

$\therefore$ ${\bf \int_{Z = e^e}^{Z = e^{e^2}} \int_{Y= e}^{Y = e^2} \int_{X = 2N + M + 1}^{\infty} (\frac{1}{Z * ln(Z) * ln(ln(Z)) * Y * ln(Y)}) \sum_{J = 1}^{N} \frac{1}{(X - J - M){(X - J - M - 1})} dx dy dz = }$

${\bf (ln(2))^3 }$

$The$ $result$ $to$ $four$ $decimal$ $places$ $is$ $.3330.$