Evaluating an Integral

JEE Mains 2015 Tutorials › Calculus questions › Evaluating an Integral

Given that
\int_{-\infty}^{\infty} e^{-x^2}\ \mathrm dx=\sqrt{\pi}
evaluate the integral
I=\int_0^\infty \dfrac{1}{\sqrt{x}}\ e^{-x}\ \mathrm dx

#Mathematics #Calculus

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6 Answers

1

Ricky ·2011-03-31 20:30:06

From ( 1 ) , we have : -

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Vivek @ Born this Way ·2011-04-01 00:54:49

Please Explain the first step ( ... Eqn (i) ) that wherefrom that 2 came. Rest is OK.

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1

Ricky ·2011-04-01 01:10:18

Let I = _{- a} âˆ« ^a f ( x ) dx = _{- a} âˆ« ⁰ f ( x ) dx + ₀ âˆ« ^a f ( x ) dx

Let J = _{- a} âˆ« ⁰ f ( x ) dx

Let " x = - z " â†’ " dx = - dz "

So , J = ₀ âˆ« ^a f ( - x ) dx .......... Since , _{- a} âˆ« ⁰ - f ( - z ) dz = ₀ âˆ« ^a f ( - z ) dz

Hence , I = ₀ âˆ« ^a f ( - x ) dx + ₀ âˆ« ^a f ( x ) dx

If " f " is an even function , then , " f ( - x ) = f ( x ) " .

Then , " I " would be = 2 ₀ âˆ« ^a f ( x ) dx

Here , " e ^{- x ²} " is an even function .

Hence , the result .

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11

Joydoot ghatak ·2011-04-01 01:12:05

@vivek, if, f(x) = f(-x)

then, _-aâˆ«^a f(x) dx =2 ₀âˆ«^a f(x).
here f(x) = f(-x).

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AKHIL ·2011-04-01 01:18:39

yup the ans is same as the given condn.

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AKHIL ·2011-04-01 01:22:07

ya coz this is an even funcn...

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