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I am looking to evaluate

$$\int_0^1 x \sinh (x) \ \mathrm{dx}$$

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    $\begingroup$ integration by parts $\endgroup$
    – Will Jagy
    Sep 21 '12 at 2:34
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Use integration by parts! General format:

$$\int f(x) g'(x)dx = f(x)g(x)-\int f'(x)g(x)dx$$

Remember! If you have bounds on the integral, they must be applied to the $f(x)g(x)$ part, too (this proceeds in the normal way).

So, if I recall correctly, the integral of $\sinh(x)$ is $\cosh(x)$...

So following the integration by parts formula, we see that

$\int_0^1 x\sinh(x)dx = (x\cosh(x)|_0^1 - \int_0^1 \cosh(x)dx$

$=\cosh(1)-[\sinh(1)-\sinh(0)] = \cosh(1) -\sinh(1) + \sinh(0)$

$=\cosh(1) - \sinh(1)$

I'm sure there's some tricky identity after that, but that's as far as I go.

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  • $\begingroup$ well not quite a tricky identity, but $\cosh 1-\sinh 1=1/e$ immediately from their definitions $\endgroup$
    – Jonathan
    Sep 21 '12 at 4:48
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You can also make use of the fact, that you can rewrite the integral:

$$\int_0^1 xf(x)\,\mathrm dx = \int_0^1 f(x) \int_0^x 1 \,\mathrm dy\,\mathrm dx.$$

This is just the integral over the right-angled triangle and using Fubini's theorem you arrive at $$\int_0^1 1 \int_y^1 f(x) \,\mathrm dx \,\mathrm dy.$$

Inserting $f(x) = \sinh(x)$ you get $$\int_0^1 x\sinh(x)\,\mathrm dx = \int_0^1 \int_y^1 \sinh(x)\,\mathrm dx\,\mathrm dy = \int_0^1 \cosh(1) - \cosh(y)\,\mathrm dy = \cosh(1) - \sinh(1).$$

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$2\sinh(x)=e^x-e^{-x}$ and $\int e^x(f(x)+f'(x))dx=e^xf(x)$

Thus, $$2\int_0^1 x \sinh (x)dx=\int_0^1 xe^xdx-\int_0^1 xe^{-x}dx$$ $$=\int_0^1 e^x(x+1)dx-\int_0^1 e^{-x}(-x+1)d(-x)-\int_0^1 e^{x}dx+\int_0^1 e^{-x}d(-x)$$ $$=xe^x|_0^1-(-xe^{-x})|_0^1-e^x|_0^1+e^{-x}|_0^1$$ $$=e+\frac{1}{e}-(e-1)+(\frac{1}{e}-1)=\frac{2}{e}$$

Hence, $$2\int_0^1 x \sinh (x)dx=\frac{2}{e}\implies \int_0^1 x \sinh (x)dx=\frac{1}{e}$$

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If you use Taylor's series representation of the integrand and then interchange the sum and integral, you simply get

$$\sum_{k=0}^{\infty} \left(\frac{1}{(2k+2)!}-\frac{1}{(2k+3)!}\right) = \frac{1}{e}.$$

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