A calculus problem by Karthik Kannan

$\displaystyle I=\int _{ 0 }^{ \infty }{ { e }^{ -({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ) }dx }$

Put $x=\frac{1}{t}$ and we have $dx = - \frac{1}{t^2} \, dt$ to get (remember to change the limits) :

$\displaystyle I = \int _{ 0 }^{ \infty }{ { \frac { 1 }{ { t }^{ 2 } } e }^{ -({ t }^{ 2 }+\frac { 1 }{ { t }^{ 2 } } ) }dt }$

Now, put $t = x$ to get

$\displaystyle I =\int _{ 0 }^{ \infty }{ { \frac { 1 }{ { x }^{ 2 } } e }^{ -({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ) }dx }$

Adding these two forms we get :

$\displaystyle 2I=\int _{ 0 }^{ \infty }{ ({ 1+\frac { 1 }{ { x }^{ 2 } } )e }^{ -({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } ) }dx }=\int _{ 0 }^{ \infty }{ ({ 1+\frac { 1 }{ { x }^{ 2 } } )e }^{ -{ (({ x }-\frac { 1 }{ { x } } ) }^{ 2 }+2) }dx }$

Put $x-\frac{1}{x} = y$ and we have $dy = ( 1+ \frac{1}{x^2} ) \, dx$ to get our integral as :

$\displaystyle 2I=\int _{-\infty }^{ \infty }{ { e }^{ -{ (y }^{ 2 }+2) }dy } ={ e }^{ -2 }\int _{ -\infty }^{ \infty }{ { e }^{ -{ y }^{ 2 } }dy }$

Now the integral left is a standard gaussian integral and it evaluates it to :

$2I={ e }^{ -2 }{ \pi }^{ \frac { 1 }{ 2 } }$

Finally $\large I=\frac { \sqrt { \pi } }{ 2{ e }^{ 2 } }$