Differentiation under integral sign

  • Thread starter neelakash
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  • #1
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Homework Statement



I have to evaluate the numerical value of the derivative of the following integral for x=1

[tex]\int_{0}^{\ ln\ x}\ e^{\ -\ x\ (\ t^2\ -\ 2)}\ dt[/tex]

Homework Equations



The formula for differentitation under integral sign.

The Attempt at a Solution



The upper limit term is straightforward:it is

[tex]\frac{\ 1}{\ x}\ e^{\ -\ x[\ (\ ln\ x)^{\ 2}\ -\ 2]}[/tex]

The other part is

[tex]\int_0^{\ ln\ x}\frac{\partial}{\partial\ x}\ e^{\ -\ x(\ t^2\ -2)}\ dt\ =\ -\ e^{\ -\ 2\ x}\ [\int_0^{\ ln\ x}\ t^2\ e^{\ -\ x\ t^2}\ dt\ -\ 2\int_0^{\ ln\ x}\ e^{\ -\ x\ t^2}\ dt\ ][/tex]

The later can be evaluated and I got the following:

[tex]\ -\ e^{\ -\ 2\ x}\ [\frac{\ -(\ ln\ x)\ e^{\ -\ x(\ ln\ x)^2}}{\ 2\ x}\ +\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{4x\sqrt{ux}}\ du\ -\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{\sqrt{ux}}\ du}][/tex]

I found the result as above.However,the two integrals neither cancel with each other nor can be evaluated.Can anyone please check and tell what should be done further.

Neel
 
Last edited:

Answers and Replies

  • #2
430
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Let,
[tex]F(x) = \int_{0}^x e^{-e^x (t^2 - 2)} \text{ d}t[/tex]
so you want to find the derivative of F(ln(x)) which you can do using the chain rule.
 
  • #3
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Does not help;it ultimately reduces to what I have got...

The thing lies in putting the limits without explicitly solving the final two inntegrals.They give zero.
 
  • #4
D H
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I helped neelakash with this problem on another forum (http://www.sciforums.com/showthread.php?t=99010). As he arrived at the correct answer there, I have no qualms posting the solution here for future reference by others.

neelakash did use the appropriate technique for differentiating under the integral sign, the Leibniz Integral Rule:

[tex]\frac{d}{dx}\int_{a(x)}^{b(x)} f(t,x)\,dt =
\int_{a(x)}^{b(x)} \frac{\partial} {\partial x} f(t,x)\,dt +
f(b(x),x)\frac{db(x)}{dx} - f(a(x),x)\frac{da(x)}{dx}[/tex]

In this particular problem,

[tex]f(t,x) = \exp\left(-x(t^2-2)\right),\quad a(x)=0, \quad b(x)=\ln x[/tex]

The partial derivative of f(t,x) wrt x is

[tex]\frac{\partial}{\partial x}f(t,x) = -(t^2-2) \exp\left(-x(t^2-2)\right)[/tex]

Applying the Leibniz Integral Rule,

[tex]\frac{d}{dx}\left(\int_0^{\ln x} \exp\left(-x(t^2-2)\right)\,dt\right) =
-\left(\int_0^{\ln x} (t^2-2) \exp\left(-x\bigl(t^2-2)\right) \,dt\right) +
\exp\left(-x(\ln^2x-2)\right)/x[/tex]


There is a sign error in the original post (that exp(-2x) should be an exp(2x)). Additionally, neelakash carried the integration a step too far. That integral on the right-hand side is evaluable in terms of the error function erf(x).


However, there is no reason to do this. neelakash finally saw the "Oh, SNAP!" light that makes this problem particularly easy. From that other forum,
neelakash said:
OK,now I think I see the crux of the matter:

whatever the indefinite integral is,the resulting form will be

[tex]\ [\ f(\ t,\ x)\ ]_{\ t\ =0}^{t\ =\ ln\ x}[/tex] at x=1

After putting t= ln x in the indefinite integral inside the square bracket and then letting x=1 is equivalent to replace x (inside the indefinite integral) by one and letting the upper limit be ln(1)

Thus,the form becomes

[tex]\ [\ ...\ ]_0^{ln(1)}\ =\ [\ ...\ ]_0^0\ =0[/tex] as the upper and lower limits are the same.OK?
 
  • #5
430
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You're right, I'm sorry for the wrong suggestion. Anyway try substituting x=1 in the terms you found. The integrals should disappear due to ln(x)=0 and the upper limit term should become e^2.
 
  • #6
HallsofIvy
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Homework Helper
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Homework Statement



I have to evaluate the numerical value of the derivative of the following integral for x=1

[tex]\int_{0}^{\ ln\ x}\ e^{\ -\ x\ (\ t^2\ -\ 2)}\ dt[/tex]

Homework Equations



The formula for differentitation under integral sign.

The Attempt at a Solution



The upper limit term is straightforward:it is

[tex]\frac{\ 1}{\ x}\ e^{\ -\ x[\ (\ ln\ x)^{\ 2}\ -\ 2]}[/tex]

The other part is

[tex]\int_0^{\ ln\ x}\frac{\partial}{\partial\ x}\ e^{\ -\ x(\ t^2\ -2)}\ dt\ =\ -\ e^{\ -\ 2\ x}\ [\int_0^{\ ln\ x}\ t^2\ e^{\ -\ x\ t^2}\ dt\ -\ 2\int_0^{\ ln\ x}\ e^{\ -\ x\ t^2}\ dt\ ][/tex]

The later can be evaluated and I got the following:

[tex]\ -\ e^{\ -\ 2\ x}\ [\frac{\ -(\ ln\ x)\ e^{\ -\ x(\ ln\ x)^2}}{\ 2\ x}\ +\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{4x\sqrt{ux}}\ du\ -\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{\sqrt{ux}}\ du}][/tex]

I found the result as above.However,the two integrals neither cancel with each other nor can be evaluated.Can anyone please check and tell what should be done further.

Neel
When x= 1, ln(x)= ln(1)= 0 so x ln(x)= 1(0)= 0. Both integrals are from 0 to 0 and so are qual to 0.
 
  • #7
511
1
Yes,we need not carry out the integral explicitly as the answer comes from observation.
 

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