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Question Number 131581 by mnjuly1970 last updated on 06/Feb/21

$$$$$$...nicecalculus...$$$$evaluate::$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x}ln\left(\frac{a+{cos}^2\left(x\right)}{b+{cos}^2\left(x\right)}\right)dx=?$$$$$$

Commented by Dwaipayan Shikari last updated on 06/Feb/21

$${\int }_0^{\mathrm{\infty }}\frac{sinxlog(a+{cos}^2\left(x\right))}{x}-\frac{sinxlog(b+{cos}^{2}x)}{x}dx$$$$={\int }_0^{\mathrm{\infty }}\frac{sinx}{x}log\left(\frac{a}{b}\right)+{\int }_0^{\mathrm{\infty }}\frac{sinx}{x}log(1+\frac{{cos}^{2}x}{a})-\frac{sinx}{x}log(1+\frac{{cos}^{2}x}{b})$$$$=\frac{\pi }{2}log\left(\frac{a}{b}\right)+\underset{z\to \mathrm{\infty }}{\lim }(f\left(z\right)-f\left(0\right))log\left(\frac{b}{a}\right)$$$$=\frac{\pi }{2}log\left(\frac{a}{b}\right)-\left(sin\left(1\right)log\left(2\right)\right)log\left(\frac{a}{b}\right)+\underset{z\to \mathrm{\infty }}{\lim }sin\left(x\right)log(1+{cos}^{2}x)$$$$=(\frac{\pi }{2}-sin\left(1\right)log\left(2\right))log\left(\frac{a}{b}\right)+..$$

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