Source

## Integration by differentiation:

Compute $\int_0^\infty xe^{-\lambda x} dx$ without integration by parts:

Consider the integral: $\int_0^\infty e^{-\lambda x} dx = \frac{1}{\lambda}$. Differentiating both side w.r.t. $\lambda$ gives:

$\frac{\partial}{\partial\lambda}\int_0^\infty e^{-\lambda x}dx = -\int_0^\infty xe^{-\lambda x}dx = -\frac{1}{\lambda^2}$

Therefore, $\int_0^\infty xe^{-\lambda x}dx = \frac{1}{\lambda^2}$.

## Integration by integration:

Compute $\int_0^1 \frac{x^b-x^a}{\ln x}dx$:

Consider $\int_0^1 x^\alpha dx = \frac{1}{\alpha + 1}$ for $\alpha > -1$. Integrate both side w.r.t. $\alpha$ from $a$ to $b$ gives: