I need to compute some Integrals for my stochastic course. And i have the following problem:
$$ \frac{\lambda^n}{\Gamma(n)} \int_0^{\infty} \exp(-\frac{\lambda}{y}) \frac{1}{y^n} dy = \star$$ so i substitute $x = \frac{1}{y}, \; dy = -\frac{dx}{x^2} $. $$\star =- \frac{\lambda^n}{\Gamma(n)} \int_0^{\infty} \exp(-\lambda x) x^{(n-1)-1} dy = -\frac{\lambda}{n-1} \int_0^{\infty} \frac{\lambda^{n-1}}{\Gamma(n-1)} \exp(-\lambda x) x^{(n-1)-1} dy = -\frac{\lambda}{n-1} $$ By the definition of the gamma distriution. But it should be $\frac{\lambda}{n-1}$. I ask myself where is my mistake?
Ok i forgot to change the boundaries, so we have: $$ \frac{\lambda^n}{\Gamma(n)} \int_0^{\infty} \exp(-\frac{\lambda}{y}) \frac{1}{y^n} dy = \star$$ so i substitute $x = \frac{1}{y}, \; dy = -\frac{dx}{x^2} $ and $\phi(y) = \frac{1}{y}$, so $\phi(\infty) = 0, \; \phi(0) = \infty$. $$\star =- \frac{\lambda^n}{\Gamma(n)} \int_\infty^{0} \exp(-\lambda x) x^{(n-1)-1} dy = \frac{\lambda}{n-1} \int_0^{\infty} \frac{\lambda^{n-1}}{\Gamma(n-1)} \exp(-\lambda x) x^{(n-1)-1} dy = \frac{\lambda}{n-1} $$