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$F(x)=\int_{-x}^{x}\frac{1-e^{-xy}}{y}dy, x>0$

I need to evaluate the above integral, I know one formula like $F(x)=\int_{v(x)}^{u(x)} f(t)dt$ then $F'(x)=(f\circ v)(x)v'(x)-(f\circ u)(x)u'(x)$, but applying this formuale I am not getting correct answer.please help

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The integrand is a function of $x$ and well as $y$. You need an extra term to deal with the integral of the partial of the integrand with respect to $x$. –  copper.hat Jan 14 '13 at 16:56

3 Answers 3

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Solution

$$F(x) = \int_{u(x)}^{v(x)} f(x,t) dt$$

$$ F'(x) = f(x,v(x))v'(x) -f(x,u(x))u'(x) + \int_{u(x)}^{v(x)} \frac{\partial f(x,t)}{\partial x}dt$$

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Hint: Use differentiation under the integral sign. Imitate one of the examples in the link. The partial derivative with respect to $x$ will get rid of the unwelcome $y$ in the denominator.

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You can write

$$\frac{1-e^{-x y}}{y} = \int_0^x dt \: e^{-t y} = x \int_0^1 dt \: e^{-t x y} $$

so that by reversing order of integration

$$ F(x) = \int_0^1 dt \: x \int_{-x}^{x} dy \: e^{-t x y} $$

or

$$ F(x) = \int_{-1}^1 dt \: \frac{\mathrm{sinh}{\: x t}}{t} $$

You may also differentiate this easily:

$$F'(x) = \int_{-1}^1 dt \: \mathrm{cosh}{\: x t} = 2 \frac{\mathrm{sinh}{\: x}}{x} $$

At this point, though, I think this is as far as you can take this particular integral.

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