How can integrate this? int_-∞ ^∞(e ^(-4t) ) dt

2 Answers
Apr 17, 2018

Diverges

Explanation:

int _(-oo) ^(oo) e^(-4t) dt

=> lim_(beta to oo) int_(-beta) ^(beta) e^(-4t) dt

=> lim_( beta to oo ) [ -1/4 e^(-4t) ]_-beta ^(beta )

=> lim_(beta to oo ) { -1/4 e^(-4beta ) - ( -1/4 e^(4beta ) ) }

=> lim_(beta to oo ) { 1/4 e^(4beta) - 1/4 e^(-4beta) }

as beta -> oo => e^(-4beta ) to 0

but beta -> oo => e^(4 beta) to oo

Hence the integral is divergent

Apr 17, 2018

The integral does not converge

Explanation:

int _(-oo) ^(oo) e^(-4t) dt = int _(-oo) ^0 e^(-4t) dt + int _0 ^(oo) e^(-4t) dt provided that both integrals converge.

int _(-oo) ^0 e^(-4t) dt = lim_(ararr-oo) int _a ^0 e^(-4t) dt

= lim_(ararr-oo)[-1/4e^(-4t)]_a^0

= lim_(ararr-oo)[-1/4 + 1/4e^(-4a)]

But lim_(ararr-oo)e^(-4a) = oo, so the integral does not converge.

Therefore, int _(-oo) ^(oo) e^(-4t) dt does not converge.