Differentiation under the integral sign

Click For Summary
Differentiation under the integral sign is a technique that allows for the introduction of a parameter into an integral, making it easier to evaluate. The method involves defining a new integral with a parameter, such as I(n, λ) = ∫₀^∞ xⁿe^(-λx) dx, where I(n, 1) equals I(n). By differentiating with respect to λ, one can relate I(n) to I(n-1), facilitating the evaluation process. This approach is similar to the algebraic principle of multiplying by 1 without changing the expression. Numerous examples and resources are available online to further illustrate this method.
Wolfxphysics12
Messages
4
Reaction score
0
I have read about this method , and how feynman utilized this method. I like doing integrals for fun, but I can't seem to understand the conceptual idea on how to introduce a parameter into the integral. Can someone , in detail, explain to me how to introduce the parameter into the integral ? (Thank you :) )
 
Physics news on Phys.org
Conceptually it is the same as multiplying a number by 1 in regular algebra: you can multiply the integrand by 1 without changing the integrand. If you are clever, it will make the integral easier to do.

i.e. we want to investigate $$I(n)=\int_0^\infty x^ne^{-x}\; dx$$

Start out by defining:
$$I(n,\lambda)=\int_0^\infty x^ne^{-\lambda x}\; dx$$ ... bearing in mind that ##I(n,\lambda)=I(n)## if ##\lambda = 1##

Now differentiate both sides by lambda.

Why may we suspect that this would help?
Because we can do I(n=0), so we only need a way to get to I(n-1) from I(n), which we can repeat until n=0 and the answer falls out, and we know what differentiating an exponential function does. This is actually what you'd do by using differentiation by parts.

There are many examples online
http://www.math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

Similar threads

  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
7K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K