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what is answer of this integral: $\int_o^{\infty}\sqrt x \sinh x dx$

(it is not in my math book) point: $\int_o^{\infty}\sqrt x \cosh x dx- \int_o^{\infty}\sqrt x \sinh x dx=\sqrt \pi$

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No wonder! It blows up. –  André Nicolas Nov 23 '12 at 19:37
    
@André Nicolas so why it gives result, take a look to the point –  Neo Nov 23 '12 at 19:54

1 Answer 1

up vote 2 down vote accepted

The integral does not converge. Express $\sinh x$ in terms of exponentials. The integral $\int_0^\infty \sqrt{x}e^x\,dx$ blows up, while $\int_0^\infty \sqrt{x}e^{-x}\,dx$ is well-behaved.

Remark: An edit to the problem seems to indicate you are interested in $\int_0^\infty \sqrt{x}(\cosh x-\sinh x)\,dx$. Expressing the hyperbolic functions in terms of exponentials, we arrive at $\int_0^\infty \sqrt{x}e^{-x}\,dx$. If you make the substitution $u=\sqrt{x}$, you will arrive at what is probably a familiar integral.

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it seems gamma function! –  Neo Nov 23 '12 at 19:56

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