Integral and Sums:?

Question

1. MORE IMPORTANT OF THE TWO:
The sum of (K^2/K!) from 1 to infinity. I know the answer (thanks Mathematica!) but not particularly sure how to get it. [2e]


2. Compute the integral from -infinity to infinity of ((e)^-x)^2. ( Hint: Let I =integral from -infinity to infinity of ((e)^-x)^2 = integral from -infinity to infinity of ((e)^-y)^2. Find I^2 by converting to polar coordinates.

Answer 1

1) If Σ k² / k! is being taken from 1 to infinity, then you rewrite this as Σ k² / k! being taken from 0 to infinity, since the first term would just be zero. Also:

Σ k² / k! =
Σ k / (k-1)! =
1/0! + Σ k / (k-1)!, where sum is from 2 onward =
1/0! + Σ (k+1) / (k)!, where sum is from 1 onward =
1/0! + Σ [(k/k!) +1/k!] =
Σ [(k/k!)] + [1/0! + Σ1/k!] =
Σ [(k/k!)] + [Σ1/k!] where the 2nd sum is now from 0 onward =
Σ [(k/k!)] + e =
Σ [1/(k-1)!] + e =
Σ [1/k!] + e (where the sum is now taken from 0 onward =
e + e = 2e.

2) I think this should work:
∫ (e^-x)^2 dx = ∫ (cos(-x) + i*sin(-x))^2 dx
∫ (e^-x)^2 dx = ∫ (cos(x) - i*sin(x))^2 dx
∫ (e^-x)^2 dx = ∫ (cos²(x) - 2i*sin(x)cos(x) - sin²(x)) dx
∫ (e^-x)^2 dx = ∫ (cos(2x) - 2i*sin(x)cos(x) ) dx
∫ (e^-x)^2 dx = (1/2)sin(2x) - i*sin²(x)
∫ (e^-x)^2 dx = (1/2)2sin(x)cos(x) - i*sin²(x)
∫ (e^-x)^2 dx = sin(x)[ cos(x) - i*sin(x) ]
∫ (e^-x)^2 dx = sin(x)[ cos(-x) + i*sin(-x) ]
∫ (e^-x)^2 dx = sin(x) (e^-x)

Answer 2

For the first, try to manipulate the series for e^x to get that one. To get you started, notice that if you differentiate the series you get:

Sum 1 to infinity k x^{k-1} /k!

This gives you one factor of k in the numerator, but you need another. How can you do that? Multiply by x and take another derivative.