I apologize in advance for my ignorance on how to type mathematical symbols in this editor.
Let F be the function defined for $x > 0$ by $F(x)= \int_1^x e^{((t^2)+1)/t}\frac{dt}{t}$ Show that $F(x) = -F(1/x)$ for all $x$ in domain of $F$.
What I have done is used the mean value theorem for integrals to say that there exists a number $z$ such that: $e^{((z^2)+1)/z} \cdot \int_1^x\frac{dt}{t} = e^{((z^2)+1)/z} \cdot (-1)\cdot\int_1^{1/x}\frac{dt}{t}$ and that they both equal $e^{((z^2)+1)/z}\ln|x|$.
What i can't figure out is how to show that the $z$ used for the first function ($F(x)$) is the same as the $z$ used in the second function ($-F(1/x)$).
Does anyone know how I could go about it?
thank you.
First, write the integral this way: $$ F(x) = \int_1^x e^{t + 1/t}\frac{dt}{t}. $$ Now we'll try the $u$-substitution $u=1/t$. Then $dt=-du/u^2$ and the new integral is $$ F(x)=\int_1^{1/x} e^{1/u+u}\cdot\frac{-du/u^2}{1/u} = -\int_1^{1/x}e^{u+1/u}\frac{du}{u} = -F(1/x). $$