Differentiation under the integral sign

Differentiation under the Integral Sign:

Differentiation under the integral sign allows us to differentiate an integral that depends on a parameter. Suppose we have an integral of the form:

$F\left(x\right)={\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x,t\right)\text{\hspace{0.17em}}dt$

Where:

• $F\left(x\right)$ is a function of $x$.
• $f\left(x,t\right)$ is a continuous function of both $x$ and $t$.
• $a\left(x\right)$ and $b\left(x\right)$ are functions of $x$ that define the limits of integration.

Theorem: If $f\left(x,t\right)$ and its partial derivative ${f}_{x}\left(x,t\right)$ are continuous on some region $R$ containing $\left(x,t\right)$ and $\frac{\mathrm{\partial }}{\mathrm{\partial }x}{\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x,t\right)\text{\hspace{0.17em}}dt$ exists, then:

$\frac{d}{dx}\left({\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x,t\right)\text{\hspace{0.17em}}dt\right)=\frac{d}{dx}F\left(x\right)={\int }_{a\left(x\right)}^{b\left(x\right)}\frac{\mathrm{\partial }}{\mathrm{\partial }x}f\left(x,t\right)\text{\hspace{0.17em}}dt+f\left(x,b\left(x\right)\right)\cdot \frac{d}{dx}b\left(x\right)-f\left(x,a\left(x\right)\right)\cdot \frac{d}{dx}a\left(x\right)$

This theorem provides a way to differentiate a function that involves a varying limit of integration with respect to a parameter $x$ by applying the derivative to both the integrand and the limits of integration.

Key points to remember:

1. Conditions for Differentiation under the Integral Sign:

• $f\left(x,t\right)$ and its partial derivative ${f}_{x}\left(x,t\right)$ should be continuous.
• The integral limits should also depend on $x$ and be differentiable.
2. Application:

• Useful in problems involving changing limits of integration, parametric integrals, or differential equations involving integrals with varying parameters.
3. Use Caution:

• The conditions for the theorem must be checked before applying it, as not all integrals permit differentiation under the integral sign.
4. Example:

• Suppose we have the integral:

$F\left(x\right)={\int }_{0}^{{x}^{2}}{e}^{xt}\text{\hspace{0.17em}}dt$

We aim to find $\frac{d}{dx}F\left(x\right)$ using differentiation under the integral sign.

Solution:

1. Given Integral: $F\left(x\right)={\int }_{0}^{{x}^{2}}{e}^{xt}\text{\hspace{0.17em}}dt$

2. Apply Differentiation under the Integral Sign:

According to the theorem, $\frac{d}{dx}\left({\int }_{a\left(x\right)}^{b\left(x\right)}f\left(x,t\right)\text{\hspace{0.17em}}dt\right)={\int }_{a\left(x\right)}^{b\left(x\right)}\frac{\mathrm{\partial }}{\mathrm{\partial }x}f\left(x,t\right)\text{\hspace{0.17em}}dt+f\left(x,b\left(x\right)\right)\cdot \frac{d}{dx}b\left(x\right)-f\left(x,a\left(x\right)\right)\cdot \frac{d}{dx}a\left(x\right)$

3. Derivative Calculation:

Let's differentiate $F\left(x\right)$ using the theorem:

$\frac{d}{dx}F\left(x\right)=\frac{d}{dx}\left({\int }_{0}^{{x}^{2}}{e}^{xt}\text{\hspace{0.17em}}dt\right)$

We differentiate with respect to $x$ within the integral:

$\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left({e}^{xt}\right)=t{e}^{xt}$

Applying the theorem:

$\frac{d}{dx}F\left(x\right)={\int }_{0}^{{x}^{2}}t{e}^{xt}\text{\hspace{0.17em}}dt+{e}^{x\cdot {x}^{2}}\cdot 2x-{e}^{x\cdot 0}\cdot 0$

4.  $\frac{d}{dx}F\left(x\right)={\int }_{0}^{{x}^{2}}t{e}^{xt}\text{\hspace{0.17em}}dt+2x{e}^{{x}^{2}}$

5. Result:

Thus, the derivative of $F\left(x\right)$ using the Differentiation under the Integral Sign theorem is: $\frac{d}{dx}F\left(x\right)={\int }_{0}^{{x}^{2}}t{e}^{xt}\text{\hspace{0.17em}}dt+2x{e}^{{x}^{2}}$