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I have a question about, how can i derivate an expression of the form $$ F\left( u \right) = \int_{C_1 }^{C_2 } {f\left( {ux} \right)\mathrm{d}x} $$ where $ C_1 ,C_2 $ are constants. I have no idea :/

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This is called differentiating under the integral sign. But since $C_1$ and $C_2$ are constants, the answer is simply $\int_{C_1}^{C_2} \frac{\partial}{\partial u}f(ux) dx$. –  Srivatsan Sep 4 '11 at 14:14
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..., which is $\int_{C_1}^{C_2}xf'(ux)\mathrm dx=\frac1u\left([xf(ux)]_{C_1}^{C_2}-\int_{C_1}^{C_2}f(ux)\mathrm dx\right)$ (where the last step is only valid for $u\neq0$). –  joriki Sep 4 '11 at 14:30
    
But for example here $$ \eqalign{ & f\left( a \right) = \int\limits_0^\infty {\frac{{Arc\tan \left( {ax} \right) - Arc\tan \left( x \right)}} {x}dx} \cr & \cr & \frac{d} {{da}}f\left( a \right) = \int\limits_0^\infty {\frac{{dx}} {{1 + a^2 x^2 }}} \cr} $$ He forgets the 1/x in the expression he only considered Arctan(ax) but I think that he should derivate with respect to a $ Arctan (ax) / x $ )= –  August Sep 4 '11 at 14:32
    
@August, The new example is not of the form you gave in your question. And the differentiation seems ok. Note that you are (partial) differentiating w.r.t. $a$, not $x$. So the $x$ is a constant. Also differentiating $\arctan (ax)$ w.r.t. $a$ gives a spare $x$, thanks to chain rule. These two $x$'s cancel nicely. –  Srivatsan Sep 4 '11 at 14:56
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up vote 3 down vote accepted

This operation is called differentiating under the integral sign. In the general case, $$ F(u) = \int_{a(u)}^{b(u)} g(u,x) dx, $$ where the limits are functions of the parameter $u$, we have $$ F'(u) = f(u, b(u)) b'(u) - f(u, a(u)) a'(u) + \int_{a(u)}^{b(u)} \frac{\partial}{\partial u} g(u,x) dx. $$ In your case, the limits are constants, and $g(u,x)$ has the special form $f(ux)$. So the first two terms drop out, giving: $$ F'(u) = \int_{C_1}^{C_2} \frac{\partial}{\partial u} f(ux) dx = \int_{C_1}^{C_2} f'(ux) x dx, $$ thanks to the chain rule.

In general, one cannot "simplify" such answers any further, but in your case you can. Noting that $$ F'(u) = \int_{C_1}^{C_2} \frac{x}{u} \frac{\partial}{\partial x} f(ux) dx, $$ and integrating by parts, we get: $$ F'(u) = \left. \frac{x}{u} f(ux) \right|_{C_1}^{C_2} - \int_{C_1}^{C_2} \left(\frac{\partial}{\partial x} \frac{x}{u} \right) \cdot f(ux) dx = \ldots $$ I will leave it to you to complete the answer. You should be able to express the final integral in terms of $F(u)$ itself.

Edit: As @joriki notes in a comment, the "simplification" step is valid only if $u \neq 0$.

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