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Let $f:R^+ \to R $ be a differentiable function such that $f(1)=3$ and satisfies $$\int_1^{xy} f(t)dt=x\int_1^{y} f(t)dt+y\int_1^{x} f(t)dt$$ $ \forall x,y\in R^+$ then find $f(e)$

My try :

Let $$F(a)=\int_1^{a} f(t)dt$$

Then our conditions give us $$F'(1)=3$$ and $$F(xy)=xF(y)+yF(x)$$

Now this form seems to be much similar to that of differentiation using product rule of two functions but this idea didn't help me much.

I tried substituting $x=y$ in above equation to get $$F(x^2)=2xF(x)$$

But I don't seem to get anywhere with this.

Rohan Shinde
  • 9,737

2 Answers2

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Taking the equation and partially differentiating with respect to $x$ gives us $$y f(xy) = \int_1^y f(t) \,\mathrm{d}t + yf(x)$$ Partially differentiating the above with respect to $y$ then gives us $$xy f'(xy) + f(xy) = f(y) + f(x)$$ Setting $y = 1$, we get $$xf'(x) + f(x) = f(x) + 3 \implies f'(x) = \frac{3}{x} \implies f(x) = 3\ln|x| + C$$ for some $C$. Clearly, from $x = 1$, we get $C = 3$, so $$f(x) = 3 \ln|x| + 3.$$ Therefore, $f(e) = 6$.

Theo Bendit
  • 50,900
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Gotcha got it.

But it is a kind of observation that from the last equation by guessing I figured out that $F(x)$ might be $3x\ln x$ and it indeed satisfies all the given conditions and hence $f(e)=6$

But still if anyone would like to given any other complete mathematical answer you are free to respond so.

Rohan Shinde
  • 9,737