Integral with a parameter

Question

Hello I try solve a problem on integration $\int_0^\frac{\pi}{2}\frac{\arctan(y\tan(x))}{\tan(x)}dx$ My stuck is that I use differentiation under integral sign and producing $ I'(y) = \int_0^ \frac{\pi}{2}\frac{dx}{1+(y\tan(x))^2}$ what can I do? Thanks

The solution is $\frac{\pi}{2}\text{sgn}(y \ln(1+|y|))$

Answer 1

First, note that the integral given by

$$I(y)=\int_0^{\pi/2}\frac{\arctan(y\tan(x))}{\tan(x)}\,dx$$

is an odd function with $I(y)=-I(-y)$. Therefore, we examine the case for $y\ge 0$.

Next, the integral given by

$$I'(y)=\int_0^{\pi/2} \frac{1}{1+y^2\tan^2(x)}\,dx \tag 1$$

can be evaluated a host of ways.

In doing so, we find that

$$I'(y)=\frac{\pi}{2(y+1)} \tag 2$$

Integrating $(1)$ and using $I(0)=0$ reveals

$$I(y)=\frac{\pi}{2}\log(1+y)$$

If $y<0$, then we use the fact that $I(y)$ is odd to write

$$I(y)=-\frac{\pi}{2}\log(1-y)$$

Putting both cases together yields

$$I(y)=\text{sgn}(y)\frac{\pi}{2}\log(1+|y|)$$

as expected!

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EVALUATION OF THE INTEGRAL IN $(1)$

We assume that $y>0$ for the proceeding analysis.

Note that we can write

$$\frac{1}{1+y^2\tan^2(x)}=\frac{1}{1-y^2}-\frac{2y^2}{1-y^2}\frac{1}{(1+y^2)+(1-y^2)\cos(2x)}$$

Then, we have

$$\begin{align} \int_0^{\pi/2}\frac{1}{1+y^2\tan^2(x)}\,dx&=\frac{\pi}{2(1-y^2)}-\frac{2y^2}{1-y^2}\int_0^{\pi/2}\frac{1}{(1+y^2)+(1-y^2)\cos(2x)}\,dx\\\\ &=\frac{\pi}{2(1-y^2)}-\frac{y^2}{1-y^2}\int_0^{\pi}\frac{1}{(1+y^2)+(1-y^2)\cos(x)}\,dx\\\\ &=\frac{\pi}{2(1-y^2)}-\frac{y^2}{2(1-y^2)}\int_{-\pi}^{\pi}\frac{1}{(1+y^2)+(1-y^2)\cos(x)}\,dx\tag 3 \end{align}$$

The integral in $(3)$ can be evaluated using the classical Tangent Half-Angle Substitution or contour integration. Here, we take the latter approach to write

$$\begin{align} \int_{-\pi}^{\pi}\frac{1}{(1+y^2)+(1-y^2)\cos(x)}\,dx&=\frac2i\oint_{|z|=1}\frac{1}{(1-y^2)z^2+2(1+y^2)z+(1-y^2)}\,dz\\\\ &=\frac{2}{i(1-y^2)}\oint_{|z|=1}\frac{1}{\left(z-\frac{y+1}{y-1}\right)\left(z-\frac{y-1}{y+1}\right)}\,dz\\\\ &=(2\pi i )\left(\frac{2}{i(1-y^2)}\right)\text{Res}\left(\frac{1}{\left(z-\frac{y+1}{y-1}\right)\left(z-\frac{y-1}{y+1}\right)}, z=\frac{y-1}{y+1}\right)\\\\ &=\frac{\pi}{y} \end{align}$$

Putting it all together we find that

$$I'(y)=\frac{\pi}{2(1+y)}$$

Answer 2

Assuming $y>0$, $$I(y)=\int_{0}^{\pi/2}\frac{\arctan(y\tan x)}{\tan x}\,dx = \int_{0}^{+\infty}\frac{\arctan(yt)}{t(1+t^2)}\,dt$$ hence: $$ I'(y) = \int_{0}^{+\infty}\frac{dt}{(1+t^2)(1+y^2 t^2)}=\frac{\pi}{2+2y} $$ by partial fraction decomposition, then $$ \forall y>0,\qquad I(y) = \int_{0}^{y}\frac{\pi\,du}{2+2u} = \frac{\pi}{2}\log(1+y) $$ and since $I(y)$ is an odd function, your claim follows.