How to Evaluate the Integral $\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{1+\sin(y)}\ln(\sin(y))}{\cos(y)}dy?$

Question

Question: How to Evaluate the Integral $$\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{1+\sin(y)}\ln(\sin(y))}{\cos(y)}dy?$$

My attempt

I'm looking for a method to evaluate it. I've attempted a substitution to simplify the expression, but I'm not sure how to proceed further. Here’s what I've done so far:

I used the substitution $\sqrt{1+\sin(y)} \rightarrow x$. This transforms the integral into:

$$-2\int_{1}^{\sqrt{2}}\frac{\ln(x^{2}-1)}{x^{2}-2}dx.$$

I further decomposed the logarithm:

$$-2\int_{1}^{\sqrt{2}}\frac{\ln(x+1)+\ln(x-1)}{(x-\sqrt{2})(x+\sqrt{2})}dx.$$

I'm stuck at this point.

Edit;The integral was given by our professor, who mentioned it has a very beautiful closed form. I have already emailed our professor, who gave the integral to my friends and me. If I receive any updates, I'll let you know.

Answer 1

You are almost at the solution. Continue with partial fraction decomposition and tou have

$$I\, \sqrt 2=-\frac{\log (x-1)}{x-\sqrt{2}}+\frac{\log(x-1)}{x+\sqrt{2}}-\frac{\log (x+1)}{x-\sqrt{2}}+\frac{\log(x+1)}{x+\sqrt{2}}$$ $$\int\frac{\log (x+a)}{x+b}\,dx=\text{Li}_2\left(\frac{a+x}{a-b}\right)+ \log (a+x) \log\left(\frac{b+x}{b-a}\right)$$

Answer 2

Starting from $$I=\int\frac{\log(x^2-1)}{x^2-2}dx$$

Notice that performing partial fraction decomposition to $\frac{\log(x^2-1)}{x^2-2}$ we get $\frac{\log(x^2-1)}{2^{3/2}(x-\sqrt{2})}-\frac{\log(x^2-1)}{2^{3/2}(x+\sqrt{2})}$ .So $$I=\frac{1}{2^{3/2}}\int\frac{\log(x^2-1)}{(x-\sqrt{2})}dx-\frac{1}{2^{3/2}}\int\frac{\log(x^2-1)}{(x+\sqrt{2})}dx$$

Let's call $I_1=\int\frac{\log(x^2-1)}{(x-\sqrt{2})}dx$ and $I_2=\int\frac{\log(x^2-1)}{(x+\sqrt{2})}dx$

For $I_1$

$$\int\frac{\log(x^2-1)}{(x-\sqrt{2})}dx$$

We can substitute $u=x-\sqrt{2}$ and $du=dx$ , so $x^2=(u+\sqrt{2})^2$

$$\int\frac{\log((u+\sqrt{2})^2-1}{u}du=\int\frac{\log(u+\sqrt2+1)+\log(u+\sqrt{2}-1)}{u}du\\\int\frac{\log(u+\sqrt{2}+1)}{u}du+\int\frac{\log(u+\sqrt{2}-1)}{u}du $$

For $\int\frac{\log(u+\sqrt{2}+1)}{u}du$ we get that

$$\int\frac{\log(\frac{u}{\sqrt{2}+1}+1)}{u}du+\log(\sqrt{2}+1)\int\frac{1}{u}du$$

Substituting $v=-\frac{u}{\sqrt{2}+1}$ and $dv=-\frac{1}{\sqrt{2} +1}du$

$$\int-\frac{\log(1-v)}{v}dv+\log(\sqrt{2}+1)\log(u)=\\-\operatorname{Li}_2(v)+\log(\sqrt{2}+1)\log(u)=-\operatorname{Li}_2(-\frac{u}{\sqrt{2}+1})+\log(\ sqrt{2} +1)\log(u)$$

Similarly for $\int\frac{\log(u+\sqrt{2}-1)}{u}du$ we get

$$\int\frac{\log(u+\sqrt{2}-1)}{u}du=-Li_2(-\frac{u}{\sqrt{2}-1})+\log(\sqrt{2}-1)\log(u)$$

Undoing the $u=x-\sqrt{2}$ substitution from the previous two results we get

$$I_1=-\operatorname{Li}_2(-\frac{x-\sqrt{2}}{\sqrt{2}+1})-\operatorname{Li}_2(-\frac{x-\sqrt{2}}{\sqrt{2}-1})+\log(\sqrt{2}+1\log(x-\sqrt{2})+\log(\sqrt{2}-1)\log(x-\sqrt{2})$$

For $I_2$

$$\int\frac{\log(x^2-1)}{x+\sqrt{2}}dx$$

Substitute $u=x+\sqrt{2}$ and $dx=du$

$$\int\frac{\log((u-\sqrt{2})^2-1)}{u}du=\int\frac{\log(u-\sqrt{2}+1)}{u}du+\int\frac{\log(u-\sqrt{2}-1)}{u}du$$

Those integrals are really similar to the ones that came from $I_1$ and they are solved with the exact same way .

We gwt

$$I_2=-\operatorname{Li}_2(-\frac{x+\sqrt{2}}{1-\sqrt{2}})-\operatorname{Li}_2(-\frac{x+\sqrt{2}}{-\sqrt{2}-1})+\log(1-\sqrt{2})\log(x+\sqrt{2}+\log{-\sqrt{2}-1})\log(x+\sqrt{2})$$

Plugging those values into the starting integral and simplifying we get

$$I=\frac{1}{2^{3/2}}I_1-\frac{1}{2^{3/2}}I_2$$

We get that $$I=\dfrac{\left(-\ln\left(1-\sqrt{2}\right)-\ln\left(-\sqrt{2}-1\right)\right)\ln\left(\left|x+\sqrt{2}\right|\right)+\left(\ln\left(\sqrt{2}+1\right)+\ln\left(\sqrt{2}-1\right)\right)\ln\left(\left|x-\sqrt{2}\right|\right)+\operatorname{Li}_2\left(\frac{x+\sqrt{2}}{\sqrt{2}+1}\right)+\operatorname{Li}_2\left(\frac{x+\sqrt{2}}{\sqrt{2}-1}\right)-\operatorname{Li}_2\left(-\frac{x-\sqrt{2}}{\sqrt{2}+1}\right)-\operatorname{Li}_2\left(-\frac{x-\sqrt{2}}{\sqrt{2}-1}\right)}{2^\frac{3}{2}}$$