66Interesting... the integration of a positive integrand (in the limit 0 to 1, the integrand is positive) turns out to be negative.
Further, check your limits after the substitution in the first step.
8 Answers
66
62I guess someone found the integral and deleted.. this is sad...
1sorry sir it got deleted by mistake
i will type it again when i have the time...
i was trying to edit it when it got deleted.....
sorry once again
the answer was ln2 if i remeber rightly
1now we know the property that
if I=∫lim a to b f(x,t)dt
then
dI/dx==∫lim a to b ∂f(x,t)/∂x dt (dont forget the partial differentiation sign)
so now let us consider a function f(x)= x^t-1/lnx dx
so let I=∫lim 0 to 1 x^t-1/lnx dx
therefore
di/dt=∫lim 0 to 1 (x^t)lnt/lnt dx
then
di/dt=∫lim 0 to 1 x^t
di/dt=x^(t+1)/t+1|lim 0 to 1
di/dt=1/t+1
di=(1/t+1)dt
now integrating both sides
i=ln|t+1| +c
now substituting t=0 we get the value of c=0
therefore
i=ln|t+1|
now in the above question t=1
so the final answer is ln|1+1|=ln2
this solution can be considered for any value of t integral or non integral
66Yes, that's right... Or one might go backwards:
For \alpha>-1
\int_0^1x^\alpha \ \mathrm dx =\dfrac{1}{\alpha +1}
Hence, integrating
\int_0^1\int_0^1x^\alpha \ \mathrm dx \mathrm d\alpha=\int_0^1\dfrac{\mathrm d\alpha}{\alpha +1}=\ln 2
Interchange the order of integration on the left (which is valid under very mild conditions)
one gets
\int_0^1\left(\int_0^1x^\alpha \ \mathrm d\alpha\right)\mathrm dx =\int_0^1\dfrac{x-1}{\ln x}\ \mathrm dx
hence the required result.