MHB Differentiating wrt constant to evaluate integral

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Differentiating integrals with respect to constants is justified by Leibniz's rule, which allows for the interchange of limits and integrals under certain conditions, such as continuity of the integrand. When treating "a" as a variable, the integrand becomes a function of two variables, enabling differentiation. The significance of differentiating with respect to a constant like "3" lies in the broader context of multivariable calculus, where "a" can be viewed as a parameter. This approach highlights that differentiation does not simply commute with evaluation. Understanding these principles is crucial for correctly applying differentiation under the integral sign.
Yashasvi Grover
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What is the justification for differentiating some integrals with respect to constants in order to obtain result, i.e. ∂/∂a[∫e^(− ax^2).dx] =∫-x^2.e^(-ax^2) dx?I mean what if we say "a" was 3 then differentiating wrt 3 would have no significance?How can we treat it like a multivariable function :/
 
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Yashasvi said:
What is the justification for differentiating some integrals with respect to constants in order to obtain result, i.e. ∂/∂a[∫e^(− ax^2).dx] =∫-x^2.e^(-ax^2) dx?I mean what if we say "a" was 3 then differentiating wrt 3 would have no significance?How can we treat it like a multivariable function :/
Regarding your first question: The theorem justifying the interchange of the limit and the integral is usually called something like "Leibniz' rule for differentiating under the integral sign". Assuming the integral limits are fixed (or at least: not functions of the differentiation variable $a$), it is sufficient that the integrand itself is continuous as a function of $x$ and $a$ and continuously differentiable w.r.t. $a$.

Regarding your second question: Nothing stops us from regarding $a$ as a variable (so the integrand indeed becomes a function of two variables). Also, differentiation does not commute with evaluation.
 

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