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Test the improper integral ∫ x / cosh x for convergence

by RoRi

Test the following improper integral for convergence:

    \[ \int_{-\infty}^{\infty} \frac{x}{\cosh x} \, dx. \]

The integral converges.

Proof. We recall the definition of the hyperbolic cosine in terms of the exponential,

    \[ \cosh x = \frac{e^x + e^{-x}}{2} = \frac{e^{2x} + 1}{2e^x}. \]

Then we have

    \begin{align*} \int_{-\infty}^{\infty} \frac{x}{\cosh x} \, dx &= 2 \int_{-\infty}^{\infty} \frac{xe^x}{e^{2x} + 1} \, dx \\[9pt] &= 2 \left( \int_{-\infty}^{-1} \frac{xe^x}{e^{2x} + 1} \, dx + \int_{-1}^1 \frac{xe^x}{e^{2x}+1} \, dx + \int_1^{\infty} \frac{xe^x}{e^{2x}+1} \, dx \right). \end{align*}

The middle integral converges since it is a proper integral of an integrable function. For the integral on the left we have

    \begin{align*} \int_{-\infty}^{-1} \frac{xe^x}{e^{2x}+1} \, dx &= \int_{\infty}^1 \frac{(-x)e^{-x}}{e^{-2x} + 1} \, dx \\[9pt] &= - \int_1^{\infty} \frac{-xe^x}{1 + e^{2x}} \, dx \\[9pt] &= \int_1^{\infty} \frac{xe^x}{e^{2x}+1} \, dx. \end{align*}

But, we have

    \[ \frac{xe^x}{e^{2x}+1} < \frac{xe^x}{e^{2x}} = \frac{x}{e^x} \]

for all x \in [1, +\infty). We know \int_1^{\infty} \frac{x}{e^x} \, dx converges by Example 4 (page 418 of Apostol); hence,

    \[ \int_1^{\infty} \frac{xe^x}{e^{2x}+1} \, dx \]

converges. Since this is also the third integral in the sum above, we have proved that every integral in the sum converges, and hence we have established the convergence of

    \[ \int_{-\infty}^{\infty} \frac{x}{\cosh x} \, dx. \qquad \blacksquare \]

3 comments

  3. Rafael Deiga says:

    When you made the substitution x=-t, you forgot a minus. Since the function x/coshx is odd, the result of integration is zero.

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