Question

Can someone help me solve this Definite Integral: Integrating variable n from 0 -> (RetirementAge-CurrentAge-1).

The formula to integrate is ((Salary*Deferral%)*(1+Inflation)^n)*(1+MarketRate)^(RetirementAge-CurrentAge-n).

For those who are curious, this is for a Retirement Finance calculation and my hope is to convert this integral into something I can put into Excel!

Answer 1

do any of the values that appear as words in this integral vary with n?

Answer 2

salary, deferral, etc?

Answer 3

Well, yes, in that n is used in the exponent of each part of the equation. Is that what you meant?

Answer 4

no, I mean that we can rteat salary,Deferral%, market rate as constant?
i.e. they are not functions of n ?

Answer 5

treat ?

Answer 6

That is mostly correct. Constants are Deferral%, Inflation, MarketRate, RetiremetnAge, and CurrentAge. Salary will be increasing though by the rate of Inflation each year. The first part of the equation, (Salary*Deferral%)*(1+Inflation)^n), represents the increase of Salary each year after the first by Inflation.

Answer 7

well mathematically we have to either know exactly how those varaibales are functions of n, or else treat them as constant....
I don't know enough about economics to say how dependent those variables are on n and whether or not we can ignore that effect

Answer 8

Treat them as constants. Only n is being integrated. Nothing else is dependent on n.

Answer 9

I see. Salary is also constant. Thus, yes, aside from n, all the rest are Constants.

Answer 10

My Calculus speak is not what it used to be...

Answer 11

if we just say all those things are constants
RetirementAge=A
Current age=a
Salary=S
Deferral%=D
Inflation=I
MarketRate=r

the integral is\[\int_0^{A-a-1}SD(1+I)^n(1+r)^{A-a-n}dn\]I guess...

Answer 12

Yes, save that I don't know what the dn is.

Answer 13

it just means that we are integrating with respect to n

Answer 14

not that I know how to do this integral offhand...

Answer 15

Yes, that is the correct formula. It's what I tried to type up front, but since this was my first post, I didn't realize i could do the fancy formatting you just did.

Answer 16

Haha, wait, you don't know how to integrate this D:< ?

Answer 17

S and D are constant, so we can tqake them out of the integral\[SD\int_0^{A-a-1}(1+I)^n(1+r)^{A-a-n}dn\]now we have two differently based exponential functions, so this is not an easy integral

Answer 18

we can even pull out (1+r)^(A-a)\[SD(1+r)^{A-a}\int_0^{A-a-1}(1+I)^n(1+r)^{-n}dn\]okay now I think I can do it :)

Answer 19

Very nice!

Answer 20

\[SD(1+r)^{A-a}\int_0^{A-a-1}(\frac{1+I}{1+r})^ndn\]it's going to be tricky to type, but this can be integrated

Answer 21

the formula to use is\[\int a^xdx=\frac{a^x}{\ln x}+C\]in your case a= that whole fraction\[\large SD(1+r)^{A-a}\int_0^{A-a-1}(\frac{1+I}{1+r})^ndn=\left .\frac{(\frac{1+I}{1+r})^n}{\ln(\frac{1+I}{1+r})}\right|_0^{A-a-1}\]

Answer 22

sorry, forgot the part I took out of the integral\[ SD(1+r)^{A-a}\int_0^{A-a-1}(\frac{1+I}{1+r})^ndn= SD(1+r)^{A-a}\left .\frac{(\frac{1+I}{1+r})^n}{\ln(\frac{1+I}{1+r})}\right|_0^{A-a-1}\]

Answer 23

so this is the same as\[{SD(1+r)^{A-a}\over \ln(\frac{1+I}{1+r})}\left .(\frac{1+I}{1+r})^n\right|_0^{A-a-1}\]

Answer 24

evaluating n this gives\[={SD(1+r)^{A-a}\over \ln(\frac{1+I}{1+r})}[(\frac{1+I}{1+r})^{A-a-1}-1]\]not sure how you want to simplify this...

Answer 25

anyway, that would be your answer if we haven't had any miscommunications along the way
hope that helped!

Answer 26

I think I can just sink this into excel lock/stock and it will work. I'll test and let you know!

Answer 27

please do!

Answer 28

Quick clarification, the part is brackets is to be multiplied by the first quotient, right?

Answer 29

right

Answer 30

\[{SD(1+r)^{A-a}\over \ln(\frac{1+I}{1+r})}\left[(\frac{1+I}{1+r})^{A-a-1}-1\right]\]here it is more clear

Answer 31

or if you prefer this is the same as\[{SD(1+r)^{A-a}\over \ln(1+I)-\ln(1+r)}\left[(\frac{1+I}{1+r})^{A-a-1}-1\right]\]there are many ways to write the answer

Answer 32

You nailed it TuringTest >:D !