How does $f$ change with the increasing of $\sigma$ The $f=\int^{+\infty}_{0}x^{\lambda}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-m)^2}{2\sigma^2}}dx$, and $\lambda$ is a positive constant value, $m=1000$, how does $f$ change with the increasing of $\sigma$. (if $\sigma^2$ changes from 20 to 200)

How to prove it? Thanks.
If the integral were taken from $-\infty$ to $\infty$, then the integral equals to $\newcommand{\expect}{\mathbf{E}} \expect\left(X^\lambda\right)$ where $X\sim\mathcal{N}(m, \sigma^2)$.

So obviously when $\lambda=1$, it is nothing but the expected value of $X$, hence it stays the same, the value is the mean, $m$.

If $\lambda=2$, then it's $\newcommand{\var}{\mathbf{Var}}\expect X^2 = \var(X) + \left(\expect X\right)^2 = \var(X) + m^2$, hence it's increasing as $\sigma$ increases, since $\var(X) = \sigma^2$.
Taking the derivative

$$f'(\sigma) = \frac{1}{\sigma^4}\int_{0}^{\infty} \frac{x^\lambda}{\sqrt{2\pi}}((x-\mu)^2 - \sigma^2)e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$$

Now let's take the case of $\mu=0$. This simplifies down to comparing the values of the two integrals 
$$I_1=\int_{0}^{\infty} x^{\lambda+2}e^{-\frac{x^2}{2\sigma^2}}dx \hspace{20 pt} I_2=\sigma^2\int_{0}^{\infty} x^{\lambda}e^{-\frac{x^2}{2\sigma^2}}dx$$

$$ I_1 = \sigma^{\lambda+3}2^{\frac{\lambda+1}{2}}\Gamma\left( \frac{\lambda+3}{2}\right) \hspace{20 pt} I_2 = \sigma^{\lambda+3}2^{\frac{\lambda-1}{2}}\Gamma\left( \frac{\lambda+1}{2}\right)$$
$$\implies \frac{I_1}{I_2} = \lambda + 1$$

In other words, this derivative is always positive for $\lambda &gt; 0$, so it is always an increasing function of $\sigma$. I'm not sure yet how this changes in the $\mu\neq 0$ case.