Definite Integration

Calculus Level 5

$\displaystyle \int _{ 0 }^{ \infty }{ { e }^{ -({ x }^{ 2 }+4x^{ -2 }) } } \ dx = \ ?$

Details and Assumptions

Does not require any advanced skills like gamma function.
Not an original, But I felt it is a nice problem

The answer is 0.01623.

2 solutions

Mvs Saketh
Mar 2, 2015

I will give a hint only, if someone requires the answer, then comment

Let the integral be 'I' replace '2' with a varriable 'a' now put a/x = z

differentiate 'I' with respect to 'a',

and you will know where to go next :)

$\displaystyle I=\int _{ 0 }^{ \infty }{ { e }^{ -({ x }^{ 2 }+\frac { 4 }{ { x }^{ 2 } } ) }dx }$

$x=\frac{2}{t}$

$\displaystyle I = \int _{ 0 }^{ \infty }{ { \frac { 2 }{ { t }^{ 2 } } e }^{ -({ t }^{ 2 }+\frac { 4 }{ { t }^{ 2 } } ) }dt } =\int _{ 0 }^{ \infty }{ { \frac { 2 }{ { x }^{ 2 } } e }^{ -({ x }^{ 2 }+\frac { 4 }{ { x }^{ 2 } } ) }dx }$

$\displaystyle 2I=\int _{ 0 }^{ \infty }{ ({ 1+\frac { 2 }{ { x }^{ 2 } } )e }^{ -({ x }^{ 2 }+\frac { 4 }{ { x }^{ 2 } } ) }dx }=\int _{ 0 }^{ \infty }{ ({ 1+\frac { 2 }{ { x }^{ 2 } } )e }^{ -{ (({ x }-\frac { 2 }{ { x } } ) }^{ 2 }+4) }dx }$

$x-\frac{2}{x} = y$

$\displaystyle 2I=\int _{-\infty }^{ \infty }{ { e }^{ -{ (y }^{ 2 }+4) }dy } ={ e }^{ -4 }\int _{ -\infty }^{ \infty }{ { e }^{ -{ y }^{ 2 } }dy }$

integral is a standard gaussian integral

$2I={ e }^{ -4}{ \sqrt{\pi} }$

$\displaystyle I=\frac { \sqrt { \pi } }{ 2{ e }^{ 4 } }$

U Z - 6 years, 3 months ago

Nice, Theres another way it can be done without using gaussian integral as well, i shall tag you after posting solution

Mvs Saketh - 6 years, 3 months ago

@Mvs Saketh can u post the solution to this problem Thanks

Deena Albert - 6 years, 3 months ago

@Mvs Saketh No there isn't. Even if we take your method of differentiating the integral we would get a differential equation in terms of a as I'(a) =-2I(a). Solving it we would have $I(a) = Ce^{-2a}$ To evaluate the constant C you would have to put a=0 and thereby you would arrive at the gaussian integral and the constant $C$ would evaluate to nothing but $\frac{\sqrt{\pi}}{2}$ . So in anyways whether you do it by @U Z method or by your method you would need the gaussian integral at some step or the other.

Arghyadeep Chatterjee - 1 year, 4 months ago

a small coorection required

it should be $x-\frac{2}{x}=y$

Tanishq Varshney - 6 years, 3 months ago

Thanks what was your method?

U Z - 6 years, 3 months ago

@U Z – I applied the same method.

Tanishq Varshney - 6 years, 3 months ago

what are gaussian integral?

Deena Albert - 6 years, 3 months ago

the integral of $e^{-x^2}$ over entire line is $\sqrt{\pi}$ . It is used for statistics, analysis and advanced calculus

Figel Ilham - 6 years, 3 months ago

@Figel Ilham – Sorry, but I think you made a mistake , it should be e^{-x^{2}} to get $\huge e^{-x^{2}}$

A Former Brilliant Member - 6 years, 3 months ago

@A Former Brilliant Member – Oh yeah. I use the brackets instead the curly one. I'll fix it.. Thanks for reminding me

Figel Ilham - 6 years, 3 months ago

HI ,I think this might be of interest to you :)

A Former Brilliant Member - 6 years, 3 months ago

I think that I did correct in waiting for this problem to get rated !!

A Former Brilliant Member - 6 years, 3 months ago

thank you for the information

Deena Albert - 6 years, 3 months ago

PLZ POST UR SOLNNNNN

rajat kharbanda - 6 years, 1 month ago

Prakhar Bindal
Sep 27, 2016

Firstly let x = t*root(2)

Then replace t by 1/t and add the two integrals

Now substitute t-1/t = y and taking somethings common you will get the pretty standard gaussian integral!

Definite Integration

The answer is 0.01623.

2 solutions

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