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Let $f:\mathbb{R^{+}}\rightarrow \mathbb{R}$ be a differentiable function with $f(1) = 3$ and satisfying::

$\displaystyle \int_{1}^{xy}f(t)dt = y\int_{1}^{x}f(t)dt+x\int_{1}^{y}f(t)dt\;\forall x,y \in \mathbb{R^{+}}\;,$ Then $f(e) = $

$\bf{My\; Try::}$ Differentiate both side w. r. to $x\;,$ we get

$\displaystyle \Rightarrow f(xy)\left(x\frac{dy}{dx}+y\right) = y\cdot f(x)+\int_{1}^{x}f(t)dt\cdot \frac{dy}{dx}+x\cdot f(y)\cdot \frac{dy}{dx}+\int_{1}^{y}f(t)dt$

Now I did not Understand How can I solve after that, Help me

Thanks

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juantheron
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1 Answers1

2

Let $F$ be an antiderivative of $f$, $F'(x)=f(x)$ for all $x\geq 1$. The given relation provides:

$$F(xy)-F(1) = y(F(x)-F(1)) + x(F(y)-F(1))$$

for all $x\geq 1, y\geq 1$.

Differentiating it with respect to $x$, we get:

$$yf(xy)=yf(x)+F(y)-F(1).$$

Differentiating this new relation with respect to $y$, we get:

$$f(xy)+yxf'(xy) = f(x)+f(y).$$

Setting in it $y=1$, we have:

$$f(x)+xf'(x)=f(x)+3$$

for all $x\geq 1$.

So:

$$xf'(x)=3,$$

which, together with $f(1)=3$, gives us:

$$f(x) = 3 \ln x +3$$

for all $x\geq 1$. In particular, $f(e)=6$.

ir7
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