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So i have this integral with parameter: $$F(y)=\int_0^{\frac{\pi}{2}} \ln{(y^2-\sin^2{x})}dx$$ also let $y>1$

So i want to define function $F(y)$ explicitly. I know i should use some partial derivatives and using properties of that multivariable function $F(x,y)$ but i don't know what to do after i define it like this: $$F(x,y)=\ln{(y^2-\sin^2{x})}$$ Any help would be appreciated.

MathIsTheWayOfLife
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  • You have defined $F$ explicitly as an integral. What exactly is it you want to do? (Note you shouldn't call both the integrand and the function defined by the integral $F$ as they are two different functions.) – smcc Aug 08 '16 at 10:11
  • @smcc it's suppose to write function without the integral sign, sorry for the confusion but it's stated : " Explicitly write function F(y)" – MathIsTheWayOfLife Aug 08 '16 at 10:38
  • Yes, but the function is defined explicitly. From your comment, it seems a more accurate statement of your problem would be: "Evaluate the integral $\int_0^{\frac{\pi}{2}} \ln{(y^2-\sin^2{x})}dx$, where $y>1$". – smcc Aug 08 '16 at 10:51
  • WolframAlpha gives the value of the integral as $\pi\ln\left(\frac{\sqrt{y^2-1}+y}{2}\right)$. (The indefinite integral looks quite messy.) – smcc Aug 08 '16 at 11:10
  • By symmetry the integral equals $\frac{1}{4}\int_0^{2\pi}\log(y^2-\sin(x)^2)dx$. Now use $z=e^{i x}$ – tired Aug 08 '16 at 11:29

1 Answers1

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Let $F(y)=\int_0^\frac{\pi}{2} f(x,y) dx$. By Leibnitz-rule, we have
$$F'(y)=\int_0^\frac{\pi}{2} f'_y(x,y)dx=\int_0^\frac{\pi}{2} \frac{2y}{y^2-\sin^2(x)}dx=\int_0^\frac{\pi}{2} \frac{2y}{y^2-1+\cos^2(x)}dx\stackrel{t=\tan{x}}=\int_0^\infty \frac{2y}{(y^2-1)t^2+y^2}=\int_0^\infty \frac{2}{y((\frac{\sqrt{y^2-1}}{y}t)^2+1)}dt\stackrel{x=\frac{\sqrt{y^2-1}}{y}t}=\frac{2}{\sqrt{y^2-1}}\int_0^\infty\frac{dx}{x^2+1}=\frac{\pi}{\sqrt{y^2-1}}$$

Therefore, $$F(y)=\int \frac{\pi}{\sqrt{y^2-1}} dy=\pi\ln\left(\frac{\sqrt{y^2-1}+y}{2}\right)+C.$$

Now, you have to calculate $F(1+)=-\pi\ln{2}$ in order to obtain $C=0$ (I leave this to you as an exercise. You can check your result here).

Finally, $$F(y)=\pi\ln\left(\frac{\sqrt{y^2-1}+y}{2}\right).$$

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