I have to evaluate the numerical value of the derivative of the following integral for x=1 \(\int_{0}^{\ ln\ x}\ e^{\ -\ x\ (\ t^2\ -\ 2)}\ dt\) The formula for differentitation under integral sign. The upper limit term is straightforward:it is \(\frac{\ 1}{\ x}\ e^{\ -\ x[\ (\ ln\ x)^{\ 2}\ -\ 2]}\) The other part is \(\int_0^{\ ln\ x}\frac{\partial}{\partial\ x}\ e^{\ -\ x(\ t^2\ -2)}\ dt\ =\ -\ e^{\ -\ 2\ x}\ [\int_0^{\ ln\ x}\ t^2\ e^{\ -\ x\ t^2}\ dt\ -\ 2\int_0^{\ ln\ x}\ e^{\ -\ x\ t^2}\ dt\ ]\) The later can be evaluated and I got the following: \(\ -\ e^{\ -\ 2\ x}\ [\frac{\ -(\ ln\ x)\ e^{\ -\ x(\ ln\ x)^2}}{\ 2\ x}\ +\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{4x\sqrt{ux}}\ du\ -\int_0^{\ x(\ ln\ x)^2}\frac{\ e^{\ -\ u}}{\sqrt{ux}}\ du}]\) 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.
Three comments: Firstly, you have a sign error. \(e^{-x (t^2-2)} = e^{2x}\,e^{-xt^2}\) You have that first factor as \(e^{-2x}\) Secondly: Those integrals most certainly can be calculated in terms of the error function, erf(x): \(\text{erf}(x) \equiv \frac 2 {\sqrt{\pi}} \int_0^xe^{t^2}dt\) Thirdly, I suspect you'll do a face-palm on this. You don't need *any* of the above. What is the definite integral of any smooth function from 0 to ln(1)?
So if I understand correctly, you want to differentiate your integral expression with respect to \(x\), and then substitute in the value \(x=1\). That should be very easy to do just by using the Fundamental theorem of calculus, i.e. \(\frac{d}{dx}\int_a^x f(t)dt=f(x)\), where \(a\) is some arbitrary constant. In this case the only extra step is to use: \(\frac{d}{dx}=\frac{d}{d(lnx)}\frac{d(lnx)}{dx}= \frac{1}{x}\frac{d}{d(lnx)}\)
Oh crap, I just realized it's not as simple as I thought, because you have \(x\) inside the integral as well as in the integration limits. I'll think about this one a little. And it looks like you already used the Fundamental Theorem, so my apologies. I think what DH said is actually correct.
neeklash did use the appropriate technique for differentiating under the integral sign, the Leibniz Integral Rule: \(\frac{d}{dx}\int_{a(x)}^{b(x)} f(t,x)\,dt = \int_{a(x)}^{b(x)} \frac{\partial} {\partial x} f(t,x)\,dt + f(b(x),x)\frac{db(x)}{dx} - f(a(x),x)\frac{da(x)}{dx}\)
In this particular problem, \(f(t,x) = \exp\left(-x(t^2-2)\right),\quad a(x)=0, \quad b(x)=\ln x\) The partial derivative of f(t,x) wrt x is \(\frac{\partial}{\partial x}f(t,x) = -(t^2-2) \exp\left(-x(t^2-2)\right)\) Applying the Leibniz Integral Rule, \(\frac{d}{dx}\left(\int_0^{\ln x} \exp\left(-x(t^2-2)\right)\,dt\right) = -\left(\int_0^{\ln x} (t^2-2) \exp\left(-x\bigl(t^2-2)\right) \,dt\right) + \exp\left(-x(\ln^2x-2)\right)/x\) Now just evaluate at x=1. It's a snap.
Hello friends,I could not be online last night;so apologies for not sending a reply. Yes,I made a typo in the sign of the exponential. But I cannot convince myself that the integral of the continuous between 0 and ln(1) should be zero or something...rather it looks correct to first evaluate the indefinite integral,then putting the limits and then,substitute x=1.
OK,now I think I see the crux of the matter: whatever the indefinite integral is,the resulting form will be \(\ [\ f(\ t,\ x)\ ]_{\ t\ =0}^{t\ =\ ln\ x}\) 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 \(\ [\ ...\ ]_0^{ln(1)}\ =\ [\ ...\ ]_0^0\ =0\) as the upper and lower limits are the same.OK?
Yes...thank you for your help...In the mean time,I tried with the suggestion of rasmhop in physicsforums;but that reduced to this case only,as it should be.
