Solve $\int_{-1}^1 f(x,t) \, dx=\log \left(\frac{t+1}{1-t}\right)$ Is it possible to find a function $f(x,t)$ so that
$$\int_{-1}^1 f(x,t) \, dx=\log \left(\frac{t+1}{1-t}\right)\text{?}$$
I tried the Laplace transform, but it did not work.
Try $f(x,t)= \frac{t}{tx+1}$. Then $\int_{-1}^1 f(x,t)=[ln(x+\frac{1}{t})]_{x=-1}^{x=1}$.
Easy!$$
\int_{-1}^1 \frac{1}{2}\log\left(\frac{t+1}{1-t}\right)dx = \log\left(\frac{t+1}{1-t}\right)
$$