A Difficult Integral

$\begin{aligned} \int_0^\infty \frac {x^4}{e^{2x-1}} & = e \int_0^\infty x^4 e^{\blue{-2}x}\ dx & \small \blue{\text{Let }k = 2} \\ & = e \int_0^\infty \frac {\partial^4}{\partial k^4} e^{-kx} dx \\ & = e \frac {\partial^4}{\partial k^4} \int_0^\infty e^{-kx} dx \\ & = e \frac {\partial^4}{\partial k^4} \left(\frac 1k \right) = \frac {4!}{k^5} e = \frac {24}{32} e = \frac 34 e \end{aligned}$

Therefore $a+b = 3+4 = \boxed 7$ .

The given integral can be written as

$\large e\int_{0}^{\infty}x^{4}e^{-2x}dx$

Substituting $2x=t$

we have

$\large \frac{e}{32}\int_{0}^{\infty}t^{4}e^{-t}dt$

So we have,

$\large \frac{e\Gamma(5)}{32}$

= $\large \frac{24e}{32}$

= $\large \frac{3e}{4}$

Integrating by parts the function $x^4e^{1-2x}$ within the given limits, we get the value of the integral as $\dfrac{3e}{4}$ , making $a=3,b=4,a+b=\boxed 7$ .