# Leibnitz Integral Rule

**(15) ** Consider a function in two variables *x* and *y*, i.e.,

\[z = f(x,y)\]

Let us consider the integral of *z* with respect to *x*, from *a* to *b*, i.e.,

\[I = \int\limits_a^b {f(x,y)dx} \]

For this integration, the variable is only *x* and not *y*. *y* is essentially a constant for the integration process. Therefore, after we have evaluated the definite integral and put in the integration limits, *y* will still remain in the expression of *I.* This means that *I* is a function of *y*.

\[ \Rightarrow \,\,\,\,\,\,\,I(y) = \int\limits_a^b {f(x,y)dx} \,\,\,\,\,\,\,\,\,...(1)\]

Our 15^{th} property says that the relation (1) can be differentiated with respect to *y* as follows:

\[\begin{align}&I'(y) = \frac{d}{{dy}}\left( {\int\limits_a^b {f(x,y)} \,dx} \right)\\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \int\limits_a^b {\frac{{\partial f(x,y)}}{{\partial y}}dx} \end{align}\]

where \(\begin{align}\frac{{\partial f(x,y)}}{{\partial y}}\end{align}\) stands for the partial derivative of \(f(x,y)\) with respect to *y*, that is , the derivative of \(f(x,y)\) w.r.t. *y*, treating *x* as a constant

Let us see the justification for this property:

\[\begin{align}&I'(y) = \mathop {\lim }\limits_{h \to 0} \frac{{I(y + h) - I(y)}}{h}\\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \mathop {\lim }\limits_{h \to 0} \frac{{\int\limits_a^b {f(x,y + h)} dx - \int\limits_a^b {f(x,y)dx} }}{h}\\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \int\limits_a^b {\left\{ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{f(x,y + h) - f(x,y)}}{h}} \right)} \right\}dx} \\\,\,\,\,\,\,\,\,\,\,\,\, &\qquad= \int\limits_a^b {\frac{{\partial f(x,y)}}{{\partial y}}dx} \end{align}\]

This property turns out to be very useful in certain cases.

**Example –22**

Evaluate \(\begin{align}I = \int\limits_0^1 {\frac{{{x^k} - 1}}{{\ln x}}dx} \end{align}\)

**Solution: ** Observe that *I* will be a function of *k*. Instead of carrying out direct integration, we use property -15:

\[\begin{align}&\frac{{dI(k)}}{{dk}} = \int\limits_0^1 {\frac{\partial }{{\partial k}}\left( {\frac{{{x^k} - 1}}{{\ln x}}} \right)dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;\;= \int\limits_0^1 {\frac{{{x^k}\ln x}}{{\ln x}}dx} \,\,\,\,\,\,\,\,\,\,\,\,\\\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;\;= \int\limits_0^1 {{x^k}dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\;\;\;= \frac{1}{{k + 1}}\end{align}\]

Thus,

\[\,dI(k) = \frac{{dk}}{{k + 1}}\]

Integrating both sides, we obtain

\[I(k) = \ln (k + 1) + C\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)\]

To obtain *C*, note from the original definition of *I* that *I *(0) = 0. Using this in (1), we obtain

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,0 = \ln 1 + C\\\, &\Rightarrow \,\,\,\,C = 0\end{align}\]

Thus,

\[I(k) = \ln (k + 1)\]

Observe again carefully the indirect route that property-15 offered us to solve this integral.

**Example –23**

Evaluate \(\begin{align}I = \int\limits_0^1 {\frac{{{x^\alpha } - {x^\beta }}}{{\ln x}}dx}\end{align}\)

**Solution: ** We again try to use property -15 to solve this integral. Let us treat *I *as a function of *a *. Therefore,

\[I(\alpha ) = \int\limits_0^1 {\frac{{{x^\alpha } - {x^\beta }}}{{\ln x}}dx} \]

Notice that

\[I(\beta ) = \int\limits_0^1 {\frac{{{x^\beta } - {x^\beta }}}{{\ln x}}dx = 0} \]

Now, using property-15 we obtain:

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\frac{{dI(\alpha )}}{{d\alpha }} = \int\limits_0^1 {\frac{\partial }{{\partial \alpha }}\left( {\frac{{{x^\alpha } - {x^\beta }}}{{\ln x}}} \right)dx} \\&\qquad\qquad\;\;\;\;= \int\limits_0^1 {\frac{{{x^\alpha }\ln x}}{{\ln x}}dx} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &\qquad\qquad\quad\;= \frac{1}{{\alpha + 1}}\end{align}\]

Thus, \(dI(\alpha ) = \frac{{d\alpha }}{{\alpha + 1}}\)

\[ \Rightarrow \,\,\,\,\,\,I(\alpha ) = \ln (\alpha + 1) + C\]

Using \(I(\beta ) = 0\) above, we obtain

\[\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,I(\beta ) = 0 = \ln (\beta + 1) + C\\ &\Rightarrow \,\,\,\,\,\,\,\,C = 0 - \ln (\beta + 1) = - \ln \left( {\beta + 1} \right)\end{align}\]

Thus,

\[\begin{align}&I(\alpha ) = \ln (\alpha + 1) - \ln (\beta + 1)\\\,\,\,\,\,\,\,\,\,\,\, &\qquad= \ln \left( {\frac{{\alpha + 1}}{{\beta + 1}}} \right)\end{align}\]

This is the required integral!

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