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## 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:

\begin{aligned} \int_a^b \int_0^1 x^\alpha dx d\alpha &= \int_a^b \frac{1}{\alpha+1} d\alpha \\ \int_0^1 \left.\frac{e^{\alpha\ln x}}{\ln x} \right|_a^b dx &= \ln\left|\frac{b+1}{a+1}\right| \\ \int_0^1 \frac{x^b - x^a}{\ln x} dx &= \ln\left|\frac{b+1}{a+1}\right| \end{aligned}