# Differentiation under the integral sign

1. Mar 9, 2014

### Wolfxphysics12

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 :) )

2. Mar 9, 2014

### Simon Bridge

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

3. Mar 9, 2014

### lurflurf

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