Beta Function
Beta Function:
Euler’s First kind integral is said to be Beta function denoted by B(m,n) \[B(m,n)=\int\limits_{0}^{1}{{{x}^{m-1}}{{\left( 1-x \right)}^{n-1}}dx}\] \[\forall m,n>0\]
Properties of Beta Function: 1) \[B(m,n)=B(n,m)\] proof by replacing \[x=1-y\] 2) \[B(m,n)=\int\limits_{0}^{\infty }{\frac{{{x}^{m-1}}}{{{\left( 1+x \right)}^{m+n}}}dx}\] proof by replacing \[x=\frac{y}{1+y}\] 3) \[B(m,n)=2\int\limits_{0}^{{}^{\pi }/{}_{2}}{{{\sin }^{2m-1}}\theta {{\cos }^{2n-1}}\theta }d\theta \] proof by replacing \[x={{\sin }^{2}}\theta \]
Gamma Function
Gamma Function:
Euler’s Second kind integral is said to be Gamma function denoted by\[\Gamma n\]
\[\Gamma n=\int\limits_{0}^{\infty }{{{x}^{n-1}}{{e}^{-x}}dx}\]
Properties of Gamma Function:1) \[\Gamma \left( n+1 \right)=n\Gamma n\] \[\forall n>0\] 2) \[\Gamma \left( 1 \right)=1\] \[\forall n\in N\] 3) \[\Gamma \left( n+1 \right)=n!\] \[\forall n\in N\] 4) \[\Gamma 0=\infty \] and \[\Gamma \left( -n \right)={{\left( -1 \right)}^{n}}\infty \]\[\forall n\in N\] 5) \[\Gamma 0=\infty \] and \[\Gamma \left( -n \right)={{\left( -1 \right)}^{n}}\infty \] \[\forall n\in N\] 6) \[\Gamma \left( \frac{1}{2} \right)=\sqrt{\pi }\] 7) \[\frac{\Gamma n}{{{a}^{n}}}=\int\limits_{0}^{\infty }{{{x}^{n-1}}{{e}^{-ax}}dx}\] proof by replacing \[x=ay\] 8) \[\Gamma n=\frac{1}{n}\int\limits_{0}^{1}{{{e}^{-{{x}^{{}^{1}/{}_{n}}}}}dx}\] proof by replacing \[x={{y}^{{}^{1}/{}_{n}}}\] 9) \[\Gamma n=\int\limits_{0}^{1}{{{\left( \log \frac{1}{x} \right)}^{n-1}}dx}\] proof by replacing \[x=\log \left( \frac{1}{y} \right)\] 10) \[\Gamma n=2\int\limits_{0}^{\infty }{{{x}^{2n-1}}{{e}^{-{{x}^{2}}}}dx}\] proof by replacing \[x={{y}^{2}}\]
Relation between Beta and Gamma Function:
\[B(m,n)=\frac{\Gamma \left( m \right)\Gamma \left( n \right)~}{\Gamma \left( m+n \right)}\] \[\forall m,n>0\]
Proof by Method 1st
Proof by Method 2ndSome Important Result:
1. \[\int\limits_{0}^{{}^{\pi }/{}_{2}}{{{\sin }^{m}}\theta {{\cos }^{n}}\theta }d\theta =\frac{\Gamma \left( \frac{m+1}{2} \right)\Gamma \left( \frac{n+1}{2} \right)~}{2\Gamma \left( \frac{m+n+1}{2} \right)}\] 2. Legendre’s Duplication Formula \[\Gamma \left( m \right)\Gamma \left( m+\frac{1}{2} \right)=\frac{\sqrt{\pi }}{{{2}^{2m-1}}}\Gamma \left( 2m \right)\] \[\forall m\in Z\] 3. \[\int\limits_{0}^{\infty }{\frac{{{x}^{n-1}}}{1+x}dx=\frac{\pi }{\sin n\pi }}\] 4. Euler’s Functional Equation \[\Gamma \left( n \right)\Gamma \left( 1-n \right)=\frac{\pi }{\sin n\pi }\] \[\forall n\in \left( 0,1 \right)\] 5. \[\int\limits_{0}^{\infty }{{{x}^{n-1}}{{e}^{-ax}}\cos bxdx}=\frac{\Gamma n}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{n/2}}}\cos \left( n{{\tan }^{-1}}\frac{b}{a} \right)\] 6. \[\int\limits_{0}^{\infty }{{{x}^{n-1}}{{e}^{-ax}}\sin bxdx}=\frac{\Gamma n}{{{\left( {{a}^{2}}+{{b}^{2}} \right)}^{n/2}}}\sin \left( n{{\tan }^{-1}}\frac{b}{a} \right)\] 7. \[\Gamma \left( \frac{1}{n} \right)\Gamma \left( \frac{2}{n} \right)\Gamma \left( \frac{3}{n} \right).....\Gamma \left( \frac{n-1}{n} \right)=\frac{{{\left( 2\pi \right)}^{{}^{n-1}/{}_{2}}}}{{{n}^{{}^{1}/{}_{2}}}}\] \[\forall n\in Z,n>1\]
Problems
Q.1 Prove that
\[B(m,n)=B(m+1,n)+B(m,n+1)\]
Q.2 Prove that
\[\int\limits_{-\infty }^{\infty }{\cos \left( \frac{\pi {{x}^{2}}}{2} \right)dx}=1\]
Q.3 Prove that
\[B(m,n)=\int\limits_{0}^{1}{\frac{{{x}^{m-1}}+{{x}^{n-1}}}{{{\left( 1+x \right)}^{m+n}}}dx}\]
Q.4 Prove that
\[\int\limits_{0}^{\infty }{\frac{{{x}^{m-1}}}{{{\left( ax+b \right)}^{m+n}}}dx}=\frac{B(m,n)}{{{a}^{m}}{{b}^{n}}}\]
Also deduce that
\[\int\limits_{0}^{\pi /2}{\frac{{{\sin }^{2m-1}}\theta {{\cos }^{2n-1}}\theta }{{{\left( a{{\sin }^{2}}\theta +b{{\cos }^{2}}\theta \right)}^{m+n}}}d\theta }=\frac{B(m,n)}{2{{a}^{m}}{{b}^{n}}}\]
Q.5 Prove that
\[\int\limits_{0}^{1}{{{x}^{n-1}}{{\left( \log \frac{1}{x} \right)}^{m-1}}dx}=\frac{\Gamma \left( m \right)}{{{n}^{m}}}\]
Q.6 Prove that
\[\int\limits_{0}^{\pi /2}{\frac{d\theta }{\sqrt{\left( a{{\sin }^{4}}\theta +b{{\cos }^{4}}\theta \right)}}}=\frac{{{\left\{ \Gamma \left( \frac{1}{4} \right) \right\}}^{2}}}{4{{\left( ab \right)}^{1/4}}\sqrt{\pi }}\]
Q.7 Prove that
\[\int\limits_{a}^{b}{{{\left( x-a \right)}^{m-1}}{{\left( b-x \right)}^{n-1}}dx}={{\left( b-a \right)}^{m+n-1}}B\left( m,n \right)\]
Q.8 Prove that
\[\int\limits_{0}^{1}{\frac{{{x}^{m-1}}{{\left( 1-x \right)}^{n-1}}}{{{\left( x+a \right)}^{m+n}}}dx}=\frac{B\left( m,n \right)}{{{a}^{n}}{{\left( 1+a \right)}^{m}}}\]
Q.9 Prove that
\[\int\limits_{0}^{1}{\frac{{{x}^{m-1}}{{\left( 1-x \right)}^{n-1}}}{{{\left( b+cx \right)}^{m+n}}}dx}=\frac{B\left( m,n \right)}{{{b}^{n}}{{\left( b+c \right)}^{m}}}\]
Differentiation and Integration under the
sign of Integration
Leibnitz’s
Law for differentiation:
If \[\]\[I=\int\limits_{a}^{b}{f(x,\alpha
)dx}\]
then
then
\[
\frac{{dI}}{{d\alpha }} = \int\limits_a^b {\frac{{\partial f}}{{\partial \alpha
}}dx} + f(b,\alpha )\frac{{\partial b}}{{\partial \alpha }} - f(a,\alpha
)\frac{{\partial a}}{{\partial \alpha }}
\]
Law for
Integration:
If \[I=\int\limits_{a}^{b}{f(x,\alpha
)dx}\]
then
integrating with respect to
taking limit form
to
\[\int\limits_{{{\alpha
}_{1}}}^{{{\alpha }_{2}}}{Id\alpha }=\int\limits_{{{\alpha }_{1}}}^{{{\alpha
}_{2}}}{\int\limits_{a}^{b}{f(x,\alpha )d\alpha
dx}}=\int\limits_{a}^{b}{\int\limits_{{{\alpha }_{1}}}^{{{\alpha
}_{2}}}{f(x,\alpha )dxd\alpha }}\]
Q.1
Evaluate \[\int\limits_{0}^{\infty
}{\frac{{{e}^{-ax}}\sin mx}{x}dx}\]
Hence deduce that \[\int\limits_{0}^{\infty
}{\frac{\sin mx}{x}dx}=\frac{\pi }{2}\]
Q.2
Show that \[\int\limits_{0}^{\infty
}{\frac{{{\tan }^{-1}}\left( ax \right)}{x\left( 1+{{x}^{2}}
\right)}dx}=\frac{\pi }{2}\log \left( 1+a \right)\]
Q.3
Evaluate \[\int\limits_{0}^{\infty
}{\frac{\log \left( 1+{{a}^{2}}{{x}^{2}} \right)}{1+{{b}^{2}}{{x}^{2}}}dx}\]
Q.4 Show that \[\int\limits_{0}^{\pi
/2}{\log \left( 1-{{e}^{2}}{{\sin }^{2}}x \right)dx}=\pi \log \left\{
\frac{1+\sqrt{\left( 1-{{e}^{2}} \right)}}{2} \right\}\]
Q.5 Show that \[\int\limits_{0}^{\infty
}{{{e}^{-{{x}^{2}}}}dx}=\frac{\sqrt{\pi }}{2}\]
Hence deduce \[\int\limits_{0}^{\infty
}{{{e}^{-a{{x}^{2}}}}{{x}^{2n}}dx}=\frac{\sqrt{\pi }}{{{a}^{n+\left( 1/2
\right)}}}\frac{1.3.5...\left( 2n-1 \right)}{{{2}^{n+1}}}\]
Q.6
Using the integral
\[\int\limits_{0}^{\pi /2}{\frac{dx}{1+\alpha \cos x}}=\frac{{{\cos
}^{-1}}\alpha }{\sqrt{\left( 1-{{\alpha }^{2}} \right)}}\] \[0\le
\alpha \le 1\]
Show that \[0\le a\le 1\], \[0\le b\le
1\]
\[\int\limits_{0}^{\pi /2}{\sec x\log \left(
\frac{1+b\cos x}{1+a\cos x} \right)}dx=\frac{1}{2}\left[ {{\left( {{\cos
}^{-1}}a \right)}^{2}}-{{\left( {{\cos }^{-1}}b \right)}^{2}} \right]\]
Deduce that
\[\int\limits_{0}^{\pi /2}{\sec x\log \left(
1+\frac{1}{2}\cos x \right)}dx=\frac{5{{\pi }^{2}}}{72}\]
Q.7
Prove that \[\int\limits_{0}^{\pi }{\frac{\log
\left( 1+\sin \alpha \cos x \right)}{\cos x}}dx=\pi \alpha \]
Q.8
Prove that \[\int\limits_{0}^{\pi /2}{\frac{\log
\left( 1+\cos \alpha \cos x \right)}{\cos x}}dx=\frac{1}{2}\left(
\frac{1}{4}{{\pi }^{2}}-{{\alpha }^{2}} \right)\]
Q.9
\[\int\limits_{0}^{a}{\frac{\log \left( 1+ax
\right)}{1+{{x}^{2}}}}dx=\frac{1}{2}\log \left( 1+{{a}^{2}} \right){{\tan
}^{-1}}a\]
Q.10
\[\int\limits_{0}^{\infty }{\frac{{{\tan
}^{-1}}ax.{{\tan }^{-1}}bx}{{{x}^{2}}}}dx=\frac{\pi }{2}\log \left\{
\frac{{{\left( a+b \right)}^{a+b}}}{{{a}^{a}}{{b}^{b}}} \right\}\]
Q.11
\[\int\limits_{0}^{\pi /2}{\log \left(
\frac{a+b\cos \theta }{a-b\cos \theta } \right)}\frac{d\theta }{\sin \theta
}=\pi {{\sin }^{-1}}\frac{b}{a}\] ,\[a>b\]
Q.12
\[\int\limits_{0}^{\pi /2}{\log \left(
{{a}^{2}}{{\cos }^{2}}\theta +{{b}^{2}}{{\sin }^{2}}\theta \right)}d\theta =\pi \log \left(
\frac{a+b}{2} \right)\] , \[a,b>0\]
Q.13
If \[\int\limits_{0}^{\infty
}{{{e}^{-ax}}}dx=\frac{1}{a}\] then prove that \[\int\limits_{0}^{\infty
}{{{e}^{-ax}}}{{x}^{n}}dx=\frac{n!}{{{a}^{n+1}}}\]
Q.14
\[\int\limits_{0}^{\infty }{{{e}^{-\left(
{{x}^{2}}+\frac{{{\alpha }^{2}}}{{{x}^{2}}} \right){{\beta
}^{2}}}}}dx=\frac{\sqrt{\pi }}{2}{{e}^{-2\alpha }}\]
Q.15
\[\int\limits_{0}^{\infty
}{{{e}^{-{{a}^{2}}{{x}^{2}}}}\cos 2bx}dx=\frac{\sqrt{\pi
}}{2a}{{e}^{-{{b}^{2}}/{{a}^{2}}}}\]
Q.16
Evaluate \[\int\limits_{0}^{\pi
}{\frac{dx}{a+b\cos x}}\] , \[a>0,\left|
b \right|<a\] Deduce that
a)
\[\int\limits_{0}^{\pi }{\frac{dx}{{{\left(
a+b\cos x \right)}^{2}}}=\frac{\pi a}{{{\left( {{a}^{2}}-{{b}^{2}}
\right)}^{3/2}}}}\]
b)
\[\int\limits_{0}^{\pi }{\frac{\cos xdx}{{{\left(
a+b\cos x \right)}^{2}}}=\frac{-\pi a}{{{\left( {{a}^{2}}-{{b}^{2}}
\right)}^{3/2}}}}\]
Q.17
Prove that \[\int\limits_{0}^{\infty
}{\frac{\cos mx}{{{a}^{2}}+{{x}^{2}}}}dx=\frac{\pi }{2a}{{e}^{-ma}}\] Hence deduce that
a)
\[\int\limits_{0}^{\infty }{\frac{x\sin
mx}{{{a}^{2}}+{{x}^{2}}}}dx=\frac{\pi }{2}{{e}^{-ma}}\]
b)
\[\int\limits_{0}^{\infty }{\frac{\sin mx}{x\left(
{{a}^{2}}+{{x}^{2}} \right)}}dx=\frac{\pi }{2{{a}^{2}}}\left( 1-{{e}^{-ma}}
\right)\]
Q.18
Prove that \[\int\limits_{0}^{1}{\frac{{{x}^{a}}-{{x}^{b}}}{\log
x}}dx=\log \frac{a+1}{b+1}\]
Q.19
Prove that \[\int\limits_{0}^{\infty }{\frac{\cos
ax-\cos bx}{x}}dx=\log \frac{b}{a}\]
Q.20
Evaluate \[\int\limits_{0}^{\infty }{\frac{\sin
rx}{x}}dx\]
Q.21 Prove that \[\int\limits_{0}^{\infty
}{\frac{{{e}^{-ax}}-{{e}^{-bx}}}{x}\cos mx}dx=\frac{1}{2}\log \left(
\frac{{{b}^{2}}+{{m}^{2}}}{{{a}^{2}}+{{m}^{2}}} \right)\]
