1. **Differentiation under the integral sign : **

*This is one of the most commonly used techniques to solve numerous numbers of questions I will be using this technique to solve many other exercises in the coming posts *

Assume that we have the following function of two variables :

$ \displaystyle \int ^{b}_a \, f(x,y) \, dx$

Then we can differentiate it with respect to y provided that f has partial continuous derivative on a chosen interval.

$ \displaystyle F'(y) = \int^{b}_{a} f_{y}(x,y)\,dx$

Now using this in many problems is not that clear you have to think a lot to get the required answer because many integral questions are just in one variable so you add the second variable and assume it is __a function of two variables__ .

**Assume we want to solve the following integral : **

$ \displaystyle \int ^{1}_{0} \, \frac{x^2-1}{\ln (x) }$

Now that seems very difficult to solve but using this technique we can solve it easily no matter how much power is x raised to . So the crux move is to decide ,where to put the second variable ! So the problem with the integral is that we have a logarithm in the denominator which makes the problem so difficult to tackle !

Remember that we can get a natural logarithm if we differentiate exponential functions i.e $F(a) = 2^a \Rightarrow \,\, F'(a) = \ln(a) \cdot 2^a$

Applying this to our problem

$ \displaystyle F(a)=\int ^{1}_{0} \, \frac{x^a-1}{\ln (x) }$

Differentiate with respect to $a$

$ \displaystyle F'(b)=\frac{d}{d a}\int^{1}_{0} \frac{x^a-1}{\ln(x)}\,\,dx$

We see that the integral is a function with two variables $f(x,a)$

So we take the partial derivative with respect to a ..

$ \displaystyle F'(a)=\int^{1}_{0} \frac{\partial }{\partial a}\left(\frac{x^a-1}{\ln(x)}\right) \,\,dx $

$ \displaystyle F'(a)=\int^{1}_{0} x^a\,\,dx$

$ \displaystyle F'(a)=\frac{x^{a+1}}{a+1} \bigl]^1_0$

$ \displaystyle F'(a)=\frac{1}{a+1}$

Now integrate wrt to a to get $ \displaystyle F(a)$

$ \displaystyle F(a)=\ln{(a+1)}+C$

$ \displaystyle F(0)=\ln{1}+C\,\,\text{hence : } C=0 $

$ \displaystyle \int^{1}_{0} \frac{x^a-1}{\ln(x)}\,\,dx=\ln{(a+1)}$

By this powerful rule we were not only able to solve the integral we also found a general formula for some $ \displaystyle a$ .

Now to solve our original integral put $ \displaystyle a=2$

$ \displaystyle \int^{1}_{0} \frac{x^2-1}{\ln(x)}\,\,dx=\ln{(2+1)}=\ln(3) $

[HW] $ \displaystyle \int^{\infty}_{0}\frac{\sin (x) }{x}\,dx$

__To Be Continued ...__