Derivative of Inverse Tangent

Let

$$L_n=\left(\prod_{k=1}^{n}\left(1+\frac{k^2}{n^2}\right)\right)^{\frac1n}$$

Take logarithm:

$$\ln L_n=\frac1n\sum_{k=1}^{n}\ln\left(1+\frac{k^2}{n^2}\right)$$

As

$$n\to\infty$$

this becomes a Riemann sum:

$$\ln L=\int_0^1\ln(1+x^2)\,dx$$

Now evaluate the integral.

Let

$$I=\int\ln(1+x^2)\,dx$$

Integrating by parts,

$$u=\ln(1+x^2),\qquad dv=dx$$

Then

$$du=\frac{2x}{1+x^2}\,dx,\qquad v=x$$

Therefore,

$$I=x\ln(1+x^2)-\int\frac{2x^2}{1+x^2}\,dx$$

Write

$$\frac{2x^2}{1+x^2}=2-\frac{2}{1+x^2}$$

Hence,

$$I=x\ln(1+x^2)-\int\left(2-\frac{2}{1+x^2}\right)dx$$

$$=x\ln(1+x^2)-2x+2\tan^{-1}x$$

Therefore,

$$\int_0^1\ln(1+x^2)\,dx=\left[x\ln(1+x^2)-2x+2\tan^{-1}x\right]_0^1$$

$$=\ln2-2+\frac{\pi}{2}$$

Thus,

$$\ln L=\ln2-2+\frac{\pi}{2}$$

Exponentiating,

$$L=e^{\ln2-2+\frac{\pi}{2}}$$

$$=2e^{\frac{\pi}{2}-2}$$

Final Answer :

$$2e^{\frac{\pi}{2}-2}$$