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Let $f(x)$ be an integrable (not necessarily differentiable) function on $[a,\infty)$. Let $g(x,y)$ be differentiable on $[a,\infty)\times [c,d)$. I can also assume that the partial derivative of $g(x,y)$ with respect to $y$ is continuous.

Define $$h(y) = \int_a^\infty g(x,y)\,f(x)\,dx.$$

When calculating the differentiation $h'(y)$ for $y\in[c,d)$, can I always put the differentiation under the integral sign?

(If not, let me know if it will work with any additional "reasonable" assumptions.)

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For example, if the derivative of $g$ with respect to $y$ is continuous, you can differentiate under the integral (it's a consequence of the dominated convergence theorem). –  Davide Giraudo Apr 1 '12 at 13:16
    
It seems that making the interval half-closed might pose a problem, since then continuity of $\frac{\partial g}{\partial y}$ does not imply that it is bounded. Then $\frac{\partial g}{\partial y} (x,y)f(x)$ may not be integrable, which is the key assumption thats typically needed to apply dominated convergence to justify differentiation under the integral sign. If $\frac{\partial g}{\partial y}$ is unbounded, you will need to do some extra work to find an integrable function which dominates the difference quotients. –  user15464 Apr 1 '12 at 13:31
    
Take a look at the question as addressed here: math.stackexchange.com/questions/12909 ; wikipedia also has some information en.wikipedia.org/wiki/Differentiation_under_the_integral_sign –  Sam Lisi Apr 1 '12 at 13:35
    
You don't need $g$ to be differentiable as a function of two variables, but rather just the existence of a $y$-partial derivative. Also, why not just write the integrand as a single function $h(x,y)$ instead of as a product $f(x)g(x,y)$? –  KCd Apr 1 '12 at 14:01
    
You can't always differentiate under the integral sign. See a counterexample in Section 10 of math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf –  KCd Apr 1 '12 at 14:03
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If $g_2(x,y)=\frac{\partial}{\partial y}g(x,y)$ is continuous in $y$, uniformly in $x$ (equicontinuous), then the Mean Value Threorem says $$ \begin{align} h(y+\Delta y)-h(y) &=\int_a^\infty(g(x,y+\Delta y)-g(x,y))f(x)\,\mathrm{d}x\\ &=\int_a^\infty\Delta y\,(g_2(x,y)+o(\Delta y))f(x)\,\mathrm{d}x\\ \end{align} $$ Therefore, $$ \begin{align} h'(y) &=\lim_{\Delta y\to0}\frac{h(y+\Delta y)-h(y)}{\Delta y}\\ &=\int_a^\infty g_2(x,y)f(x)\,\mathrm{d}x+\lim_{\Delta y\to0}\frac{o(\Delta y)}{\Delta y}\int_a^\infty f(x)\,\mathrm{d}x\\ &=\int_a^\infty\frac{\partial}{\partial y}g(x,y)f(x)\,\mathrm{d}x \end{align} $$

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