Where am I going wrong: find $T$ such that compound interest equals force interest Fund A accumulates at a rate of $3 \%$ convertible monthly. Fund B accumulates with force of interest of $\delta_t=t^{20} $for all t. At time $t = 0$, $\$ 1$ is deposited in each fund. The positive time, in years, that the two funds are equal is denoted by $T$. Calculate $T$.

My attempt:
$\left(1+\frac{0.03}{12}\right)^{12t}=e^{\int _0^t\frac{t}{20}dt}-1$
$1.0025^{12t}=e^{\frac{t^2}{40}}-1$
I cannot solve this equation any further and am not confident that my formula is correct. Any help would be great!
 A: I assume that $\delta_t=\frac{t}{20}$ Then the accumulated amounts are

*

*monthly compounded interest: $1.0025^{12t}$


*forced interest: $\large{e^{\frac{t^2}{40}}}$ (without -1)
Next we make a transformation: $1.0025^{12t}=\left(\large{e^{\ln(1.0025)}}\right)^{{12t}}\normalsize =\large{e^{\ln(1.0025)\cdot 12t}}$
Next we set the exponents equal:
$$\ln(1.0025)\cdot 12t=\frac{t^2}{40}$$
Dividing the eqution by $t$, assuming that $t\neq 0$
$$\ln(1.0025)\cdot 12=\frac{t}{40}\Rightarrow t^*=T=\ln(1.0025)\cdot 12\cdot 40=1.198...\approx 1.2$$
Remark:
$0.2$ years is equal to $2.4$ months. The $0.4$ months are usually not proportionately compounded monthly. I cannot say if its play a role here.
A: It is unclear from your work whether you intend $$\delta_t = t^{20},$$ or $$\delta_t = \frac{t}{20}.$$  The question posits the former but your work uses the latter.
That said, the accumulated value of Fund B from time $t = 0$ to time $t = T$ is actually $$\exp \int_{t=0}^T \delta_t \, dt.$$ You do not subtract $1$.  Thus, for the two funds to have equal accumulated value at time $T$, we would require $$\left(1 + \frac{i^{(12)}}{12}\right)^{12T} = \exp \int_{t=0}^T \delta_t \, dt.$$  If $i^{(12)} = 0.03$, the nominal monthly rate of interest, and we assume $\delta_t = t^{20}$, then this amounts to solving $$\left(\frac{401}{400}\right)^{12T} = e^{T^{21}/21}.$$  Taking logarithms yields $$12T \log \frac{401}{400} = \frac{T^{21}}{21}.$$  Assuming $T \ne 0$, we divide by $T$ and rearrange terms to obtain $$T^{20} = 21 \cdot 12 \log \frac{401}{400},$$ hence $$T = \left( 252 \log \frac{401}{400} \right)^{1/20} \approx 0.977102.$$  This is the unique positive solution and represents just under a year.
If $\delta_t = t/20$, then we would instead solve $$\left(\frac{401}{400}\right)^{12T} = e^{T^2/40},$$ and the same method as above applies, yielding $$T = 480 \log \frac{401}{400} \approx 1.1985.$$
Note that both solutions assume that although Fund A's is nominally compounded at a monthly rate, it is effectively compounded continuously at a constant force of interest equivalent to $i^{(12)} = 0.03$, rather than discretely each month.  Thus, fractional months accrue interest for Fund A.
If we do not make this assumption, then the solution becomes slightly more complex.  If we assume interest in Fund A accrues in discrete steps at the end of each month, then the equation of value changes to $$\left(1 + \frac{i^{(12)}}{12}\right)^{12 \lfloor T \rfloor} = \exp \int_{t=0}^T \delta_t \, dt,$$ where $\lfloor T \rfloor$ is the greatest integer less than or equal to $T$.  Solving this equation requires a numerical approach.  We can use a computer to get $$T \approx 1.09476$$ for the second case.  The first case lacks a positive solution because for $0 \le T < 1$, $\lfloor T \rfloor = 0$, which implies $T = 0$.  If $1 \le T < 2$, then $\lfloor T \rfloor = 1$, but $$\left(\frac{401}{400}\right)^{12} < e^{1/21}$$ for all such $T$, and the inequality only gets more severe as $T \ge 2$.
