The equation is $$ \int_{-\infty}^\infty \exp(4t − \exp(t))\ dt $$ I am not sure what variable should I change here? Does $u=\exp(t)-4t$ work?
how to reduce the integral to a Gamma function
0
$\begingroup$
calculus
integration
gamma-function
2 Answers
6
Hint: $e^t=u$ then $$ \int_{-\infty}^\infty \exp(4t − \exp(t))\ dt = \int_{0}^\infty u^3e^{-u} dt =\Gamma(4) =6 $$
2
Hint:
Just substitute $u = e^t $ and we get $$I = \int_{0}^{\infty} u^3e^{-u} \mathrm {d}u = \Gamma (3+1) = \Gamma (4) =6$$ Hope it helps.