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How do you evaluate the integral #int e^(-sqrtx)/sqrtx# from 0 to #oo#?

Answer 1

Cameron Alexander

#2#

start with the observation that #d/dx (e^(f(x)) ) = f'(x) e^(f(x))#

and so #d/dx (e^( -sqrt x)) = - 1/(2 sqrt x) e^(- sqrt x)#

# int_0^oo e^(-sqrtx)/sqrtx dx #

#= -2 int_0^oo - e^(-sqrtx)/(2 sqrtx) dx #

#= -2 int_0^oo d/dx (e^( -sqrt x)) dx#

#= -2 [ e^( -sqrt x) ]_0^oo#

#= -2 (0 -1) = 2#

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Answer 2

Charles Anctil

To evaluate the integral ∫ e^(-√x)/√x from 0 to ∞, we first notice that the integrand has a singularity at x = 0. We can rewrite the integral as a limit of a definite integral as follows:

∫ e^(-√x)/√x dx = lim (t→0+) ∫ e^(-√x)/√x dx from t to ∞

Next, we can make a substitution to simplify the integral. Let u = √x, then du = (1/2√x) dx. This gives us:

∫ e^(-√x)/√x dx = 2∫ e^(-u) du from √t to ∞

Now we can integrate e^(-u) with respect to u:

= -2e^(-u) from √t to ∞

= -2(e^(-∞) - e^(-√t))

Since e^(-∞) = 0, the integral simplifies to:

= 2e^(-√t)

Taking the limit as t approaches 0+:

lim (t→0+) 2e^(-√t) = 2e^(-0) = 2

Therefore, the value of the integral ∫ e^(-√x)/√x from 0 to ∞ is 2.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.