Integrals are just infinite sums
Given that the integral above can be expressed as where and are integer constants between and to be found, compute
The answer is 0.1.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Note: e x ≡ n = 0 ∑ ∞ n ! x n I = ∫ − s s e − α x t d x = ∫ − s s n = 0 ∑ ∞ n ! ( − α x t ) n d x = n = 0 ∑ ∞ n ! ( − 1 ) n α n ∫ − s s x n t d x = n = 0 ∑ ∞ n ! ( − 1 ) n α n ( n t + 1 x n t + 1 ∣ ∣ ∣ ∣ − s s ) = n = 0 ∑ ∞ n ! ( n t + 1 ) ( − 1 ) n α n ( s n t + 1 − ( − 1 ) n t + 1 s n t + 1 ) = n = 0 ∑ ∞ n ! ( n t + 1 ) ( − 1 ) n α n s n t + 1 ( 1 − ( − 1 ) n t + 1 ) Since − 1 ⋅ ( − 1 ) n t + 1 = − 1 ⋅ − 1 ⋅ ( − 1 ) n t = ( − 1 ) n t : I = n = 0 ∑ ∞ n ! ( n t + 1 ) ( − 1 ) n α n s n t + 1 ( 1 + ( − 1 ) n t ) ∴ C 1 = − 1 , C 2 = 1 ⟹ C 2 1 0 C 1 = 1 0 − 1 = 0 . 1