For the following expression could someone provide some help/suggestions of my progress please
$g(x) = \int_{-ln(x)}^{x^2} \phi(t+x) dt $
The question asks me to evaluate $g'(1)$ in terms of $\phi$.
I started by finding the pdf of $\phi$ as the derivative of the CDF is the pdf. So I got the following:
$g'(x) = \int_{-ln(x)}^{x^2} \frac{-(t+x)}{\sqrt{2\pi}}\exp(-\frac{(t+x)^2}{2}) dt $
Evaluated at $x =1$ yields :
$ \phi'(1) = \frac{-(t+1)}{\sqrt{2\pi}}\exp(-\frac{(t+1)^2}{2}) $
at $x=1$ the bounds of the integral become between 0 and 1
So I think we have the following:
$g'(1) = \int_{0}^{1} (t+1)\phi'(t+1) dt = \int_{0}^{1} \frac{-(t+1)}{\sqrt{2\pi}}\exp(-\frac{(t+1)^2}{2}) dt $
But I am pretty sure this is wrong. If someone could guide me on this I would be very grateful.
Link to question: http://www.math.kent.edu/~oana/math60070/InterviewProblems.pdf
One easy way to find $g'$ is to write $g(x)=\int_{x-\ln x}^{x+x^{2}} \phi (s)ds$ which gives $g'(x)=\phi(x+x^{2})(1+2x)-\phi(x-\ln x) (1-\frac 1 x)$. So $g'(1)=3\phi (2)$.