Proof that INT(1/x)dx = lnxDate: 11/08/96 at 22:14:34 From: Ben Faulkner Subject: Integrating (1/x)dx My friend and I would like to know how to integrate (1/x)dx. Our teacher just told us that we couldn't, or at least not yet. Is there a way? Date: 11/09/96 at 08:17:59 From: Doctor Anthony Subject: Re: Integrating (1/x)dx Dear Ben and Friend, The integral of 1/x is intimately bound up with the exponential and log functions. Some people like to start with the log function and get an expression for the exponential function, or, and I prefer this method, begin with exponential functions and then move to the log function. Exponential growth can be thought of as compound interest growth, with the intervals of adding interest getting shorter and shorter until it is continuous growth - as would be the case for growth of a plant or animal. The formula for compound interest is A = P(1 + r/100)^t where A = amount, P = principal, r = rate percent per annum and t = time in years. If interest were added twice yearly, this formula would become: A = P(1 + r/200)^(2t) If interest were added s times a year: A = P(1 + r/(100s))^st Now put r/(100s) = 1/n, so s = nr/100: A = P(1 + 1/n)^(nrt/100) A = P[(1 + 1/n)^n]^(rt/100) Now if we let n -> infinity so that interest is added continuously, the expression in [ ] brackets (1 + 1/n)^n is denoted by e (by definition), and we have: A = Pe^(rt/100) We can evaluate e using the binomial theorem. If you want to learn more about the binomial theorem, take a look at: http://mathforum.org/dr.math/problems/fama.7.10.96.html Continuing on with evaluating (1 + 1/n)^n: (1 + 1/n)^n = 1 + n(1/n) + (n(n-1)/2!)(1/n^2) + (n(n-1)(n-2)/3!)(1/n^ 3) + .. Dividing the n's on the bottom line into the brackets we get: = 1 + 1 + 1(1-1/n)/2! + 1(1-1/n)(1-2/n)/3! + ... As n -> infinity: = 1 + 1 + 1/2! + 1/3! +..... to infinity This series gives e = 2.71828183 to 8 places, but e in decimal form is an unending, non-repeating, irrational number like pi or sqrt(3). If you now consider (1 + 1/n)^(nx) = e^x and expand again by the binomial theorem you get: e^x = 1 + x + x^2/2! + x^3/3! + ... to infinity. Now comes the very important part. If you differentiate e^x you get d(e^x)/dx = 0 + 1 + 2x/2! + 3x^2/3! + etc = 1 + x + x^2/2! + etc = e^x (back to where we started) So the function e^x has the UNIQUE property that it is equal to its differential coefficient. So if y = e^x then dy/dx = e^x = y dy/dx = y Then dy/y = dx and integrating INT[dy/y] = x + const. So to solve this integral dy/y in terms of y we need to express x in terms of y. We had e^x = y, so we take the natural logarithm (which means logs to base e and is denoted by "ln") of each side and we have: x = ln(y) Now we see how to integrate dy/y. We have INT[dy/y] = ln(y) + const. This would be the expression whatever letter is the variable. So if x is the variable we can write: INT[dx/x] = ln(x) + const. As you can see it is quite a long story, but it is a watershed in advancing to new branches of mathematics, with greatly improved applications to real world problems. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/