$$f(x)=\sin(x)+\int_{-\pi/2}^{\pi/2}\left(\sin(x)+t\cos(x)\right)f(t)\,\mathrm dt$$

Find maximum and minimum values of $f(x)$.

I tried to simplify this expression by checking even or odd property of $f(x)$.

We can write the above expression as



$$I_1=\int_{-\pi/2}^{\pi/2}f(t)\,\mathrm dt$$


$$I_2=\int_{-\pi/2}^{\pi/2}tf(t)\,\mathrm dt$$

if $f$ is even $I_2=0$ and if $f$ is odd $I_1=0$, but if $f$ is even

$$f(x)=(1+I_1)\sin(x)$$ which is odd. Similarly if $f$ is odd


which is neither even nor odd.

So $f(x)$ is neither even nor odd. Now to find maxima or minima $f'(x)=0$ i.e.,

$$f'(x)=(1+I_1)\cos(x)-I_2\sin(x)=0$$ $\implies$


Unable to proceed further...

  • $\begingroup$ Isn't it a Fred-Holm of second kind? $\endgroup$ – mrs Jun 7 '13 at 7:16

To compute $I_1, I_2$:

We know that $f$ satisfies

$$ (*) f(x) = (1 + I_1) \sin x + I_2 \cos x $$

We integrate! First integrate both sides of $(*)$ with respect to $x$ over $[-\pi/2,\pi/2]$. The left hand side gives $I_1$. The first term of the right hand side is odd and so vanishes under the integral, and we obtain the relation $$ I_1 = I_2 \int_{-\pi/2}^{\pi/2} \cos x dx $$ Second, hit both sides of $(*)$ with a factor of $x$ and integrate over the same interval. $x \cos x$ is odd and it vanishes under the integral; the left hand side gives $I_2$, so we have $$ I_2 = (1 + I_1) \int_{-\pi/2}^{\pi/2} x \sin x dx $$ With these two relations, you can compute $I_1, I_2$ with some simple manipulations. Now use your formula for critical points and you'll have found your extrema.

  • $\begingroup$ Very nice Thank you.. $\endgroup$ – Umesh shankar Jun 7 '13 at 8:15

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