Starting with #(epsilonA_(s,r)sigma)/(rhoVc)int_(0)^(t)dt = int_(T_i)^(T) 1/(T_(sur)^4 - T^4)dT#, show that . . . ?

LEFT-HAND INTEGRAL

The left-hand integral is trivial, as there are a lot of constants:

#(epsilonA_(s,r)sigma)/(rhoVc)int_(0)^(t)dt = (epsilonA_(s,r)sigma)/(rhoVc) cdot t#

RIGHT-HAND INTEGRAL: PREPARATION

The right-hand integral starts as:

#int_(T_i)^(T) 1/(T_(sur)^4 - T'^4)dT'#

We wrote #T'# here because it need not be equal to #T#. This is clearly a difference of quartics, so one can factor as follows:

#1/(T_(sur)^4 - T'^4) = 1/((T_(sur)^2 + T'^2)(T_(sur)^2 - T'^2))#

#= 1/((T_(sur)^2 + T'^2)(T_(sur) + T')(T_(sur)-T'))#

This becomes a partial fraction decomposition. We propose, at a constant surrounding temperature, the following two integrands:

#1/((T_(sur)^2 + T'^2)(T_(sur) + T')(T_(sur)-T'))#

#= overbrace((AT' + B)/((T_(sur)^2 + T'^2)))^"Integrand 1" + overbrace(C/(T_(sur) + T') + D/(T_(sur) - T'))^"Integrand 2"#

RIGHT-HAND INTEGRAL: ACQUIRING SYSTEM OF EQNS

When getting common denominators, one obtains:

#1 = (AT' + B)(T_(sur)^2 - T'^2) + C(T_(sur)^2 + T'^2)(T_(sur) - T') + D(T_(sur)^2 + T'^2)(T_(sur) + T')#

Next, simplify by distributing and re-factoring.

#1 = color(red)(AT_(sur)^2T') - color(orange)(AT'^3) + color(purple)(BT_(sur)^2) - color(cyan)(BT'^2)#

#+ color(purple)(CT_(sur)^3) - color(red)(CT_(sur)^2T') + color(cyan)(CT_(sur)T'^2) - color(orange)(CT'^3)#

#+ color(purple)(DT_(sur)^3) + color(red)(DT_(sur)^2T') + color(cyan)(DT_(sur)T'^2) + color(orange)(DT'^3)#

This regroups into a polynomial in orders of #T'# up to #3#:

#0T'^3 + 0T'^2 + 0T' + 1#

#= color(orange)((-A - C + D))T'^3 + color(cyan)((-B + CT_(sur) + DT_(sur)))T'^2 + color(red)((AT_(sur)^2 - CT_(sur)^2 + DT_(sur)^2))T' + color(purple)((BT_(sur)^2 + CT_(sur)^3 + DT_(sur)^3))#

As a result, we have the following system of equations (since #T_(sur) ne 0#, we divided it out of #(3)#):

#-A - C + D = 0# #" "" "" "" "" "bb((1))#
#-B + CT_(sur) + DT_(sur) = 0# #" "" "bb((2))#
#A - C + D = 0# #" "" "" "" "" "" "bb((3))#
#BT_(sur)^2 + CT_(sur)^3 + DT_(sur)^3 = 1# #" "bb((4))#

RIGHT-HAND INTEGRAL: SOLVING SYSTEM OF EQNS

In solving this we could do the following.

• Since #-A - C + D = A - C + D#, it follows that #A = -A#, which is only true if #color(green)(A = 0)#.
• Therefore, from #(1)# or #(3)#, #C = D#, and we can solve for #B# in #(2)#.

#=> B = 2CT_(sur) = 2DT_(sur)#

• To solve for #C# or #D#, use #(4)#:

#2CT_(sur) cdot T_(sur)^2 + CT_(sur)^3 + CT_(sur)^3 = 1#

#=> color(green)(C = 1/(4T_(sur)^3) = D)#

• Therefore, #color(green)(B) = 2 cdot 1/(4T_(sur)^3) cdot T_(sur) = color(green)(1/(2T_(sur)^2))#.

RIGHT-HAND INTEGRAL: SETTING IT UP

This means we can insert #A#, #B#, #C#, and #D# to get:

#1/((T_(sur)^2 + T'^2)(T_(sur) + T')(T_(sur)-T'))#

#= color(green)((1//2T_(sur)^2))/((T_(sur)^2 + T'^2)) + color(green)((1//4T_(sur)^3))/(T_(sur) + T') + color(green)((1//4T_(sur)^3))/(T_(sur) - T')#

#= overbrace(1/2 1/T_(sur)^2 1/((T_(sur)^2 + T'^2)))^"Integrand 1" + overbrace(1/4 1/T_(sur)^3 [1/(T_(sur) + T') - 1/(T_(sur) - T')])^"Integrand 2"#

The integration is now of the following:

#int_(T_i)^(T) 1/(T_(sur)^4 - T'^4)dT'#

#= overbrace(1/2 1/T_(sur)^2 int_(T_i)^(T) 1/(T_(sur)^2 + T'^2)dT')^"Integral 1" + overbrace(1/4 1/T_(sur)^3 int_(T_i)^(T) [1/(T_(sur) + T') - 1/(T_(sur) - T')]dT')^"Integral 2"#

RIGHT-HAND INTEGRAL: SOLVING IT

The second integral here is (before evaluating the integral bounds):

#color(highlight)(1/4 1/T_(sur)^3 int[1/(T_(sur) + T') - 1/(T_(sur) - T')]dT')#

#= 1/(4T_(sur)^3) [ln|T_(sur) + T'| - ln|T_(sur) - T'|]#

#= color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T')/(T_(sur) - T')|)#

The first integral here needs to be manipulated more. Factor out a #T_(sur)^2# to get:

#color(darkblue)(1/(2T_(sur)^2) int1/(T_(sur)^2 + T'^2)dT')#

#= 1/(2T_(sur)^4) int1/(1 + (T'//T_(sur))^2)dT'#

This resembles the integral of #arctanu#. Let #u = T'//T_(sur)#, so that #du = dT'//T_(sur)#.

Thus, #T_(sur)du = dT'#, and that then gives us the cubed term on the outside like we expected when we integrate:

#=> 1/(2T_(sur)^3) int1/(1 + u^2)du#

#= 1/(2T_(sur)^3) arctanu#

#= color(darkblue)(1/(2T_(sur)^3) arctan((T')/T_(sur)))#

FORMING/SIMPLIFYING THE RESULT

Now we can combine the two overarching integrals to get:

#int_(T_i)^(T) 1/(T_(sur)^4 - T'^4)dT'#

#= {:[color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T')/(T_(sur) - T')|) + color(darkblue)(1/(2T_(sur)^3) arctan((T')/T_(sur)))]|:}_(T_i)^(T)#

Evaluating this on the integral bounds, we get:

#= [color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T)/(T_(sur) - T)|) + color(darkblue)(1/(2T_(sur)^3) arctan((T)/T_(sur)))] - [color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T_i)/(T_(sur) - T_i)|) + color(darkblue)(1/(2T_(sur)^3) arctan((T_i)/T_(sur)))]#

Regroup terms to get:

#= color(highlight)(1/(4T_(sur)^3) ln|(T_(sur) + T)/(T_(sur) - T)| - 1/(4T_(sur)^3) ln|(T_(sur) + T_i)/(T_(sur) - T_i)|) + color(darkblue)(1/(2T_(sur)^3) arctan((T)/T_(sur)) - 1/(2T_(sur)^3) arctan((T_i)/T_(sur)))#

To factor this, multiply the third and fourth terms by #2/2#, then factor out #1/(4T_(sur)^3)# to get:

#= 1/(4T_(sur)^3) {color(highlight)(ln|(T_(sur) + T)/(T_(sur) - T)| - ln|(T_(sur) + T_i)/(T_(sur) - T_i)|) + color(darkblue)(2[arctan((T)/T_(sur)) - arctan((T_i)/T_(sur))])}#

FINALIZING THE RESULT

Now, we can equate this entire right-hand integral with the left-hand result we got on the first few lines:

#(epsilonA_(s,r)sigma)/(rhoVc) cdot t#

#= 1/(4T_(sur)^3) {ln|(T_(sur) + T)/(T_(sur) - T)| - ln|(T_(sur) + T_i)/(T_(sur) - T_i)| + 2[arctan((T)/T_(sur)) - arctan((T_i)/T_(sur))]}#

Multiply the left-hand constants over to get:

#color(blue)(t = (rhoVc)/(4epsilonA_(s,r)sigmaT_(sur)^3) {ln|(T_(sur) + T)/(T_(sur) - T)| - ln|(T_(sur) + T_i)/(T_(sur) - T_i)| + 2[arctan((T)/T_(sur)) - arctan((T_i)/T_(sur))]})#