I want to solve these integrals $\int \limits_{0}^{\infty}\frac{1}{1+x^a} dx $ where $a>1$ and $\int \limits_{0}^{\infty} e^{-a x^2}\cos(b x) dx$ where $a,b>0$,

I saw them in a book as exercises and i saw the answers but i am interested in the way of solving which the book does not present, it simply says using complex function integration methods as P.V., but i want simpler method if there is, i don't want all the heavy machinery from complex functions

I did try some substitutions and integration by parts but nothing seems to work !

Thanks in advance.

  • 2
    $\begingroup$ i doubt you can find a simple way $\endgroup$ – mathworker21 Mar 4 at 0:19
  • 2
    $\begingroup$ The first integral can be solved using the Beta Function. $\endgroup$ – aleden Mar 4 at 0:37
  • $\begingroup$ The second one can be gotten from the Euler one, $\int_{-\infty}^\infty e^{-x^2} dx$, by writing the cosine as an exponential, completing the square and doing some substitutions (as long as you don't mind imaginary numbers in the substitution). $\endgroup$ – NickD Mar 4 at 0:47

There are multiple approaches to each of these integrals (which are covered on this page).

For the first integral, I cover it using the Beta function here.

For the second integral, two methods that jump out are (1) Differential Under the Curve (2) Laplace Transforms. Let us consider each

For (1): We first observe that: \begin{equation} I(a,0) = \int_0^\infty e^{-ax^2}\:dx = \frac{1}{2}\sqrt{\frac{\pi}{a}} \end{equation} We proceed with $I(a,b)$ by employing Leibniz's Integral rule and taking the derivative with respect to $b$: \begin{align} \frac{\partial I}{\partial b} = \int_0^\infty e^{-ax^2} \cdot -x\sin(bx)\:dx = \int_0^\infty -xe^{-ax^2} \sin(bx)\:dx \nonumber \end{align} Here we employ integration by parts: \begin{align} u(x) &= \sin(bx) & v'(x)&= -xe^{-ax^2} \nonumber \\ u'(x) &= b\cos(bx) & v(x) &= \frac{1}{2a}e^{-ax^2} \nonumber \end{align} Thus, \begin{align} \frac{\partial I}{\partial b} &= \left[\sin(bx) \cdot \frac{1}{2a}e^{-ax^2} \right]_0^\infty - \int_0^\infty \frac{1}{2a}e^{-ax^2} \cdot b\cos(bx)\:dx \nonumber \\ &= -\frac{b}{2a}\int_0^\infty e^{-ax^2}\cos(bx)\:dx = -\frac{b}{2a}I(a,b) \end{align} Thus we have formed a differential equation with respect to $b$: \begin{equation} \frac{\partial I}{\partial b} + \frac{b}{2a}I(a,b) = 0 \end{equation} Here we employ the Integrating Factor: \begin{equation} \frac{\partial }{\partial b} \left[e^{\frac{b^2}{4a}}I(a,b) \right] = 0 \Longrightarrow e^{\frac{b^2}{4a}}I(a,b) = C \Longrightarrow I(a,b) = Ce^{-\frac{b^2}{4a}} \end{equation} Where $C$ is the constant of integration. We resolve $C$ using the condition $I(a,0)$ as given before: \begin{equation} I(a,0) = \frac{1}{2}\sqrt{\frac{\pi}{a}} = C \cdot e^{-\frac{0^2}{4a}} \Longrightarrow C = \frac{1}{2}\sqrt{\frac{\pi}{a}} \end{equation} Thus, we arrive at: \begin{equation} I(a,b) = \frac{1}{2}\sqrt{\frac{\pi}{a}} e^{-\frac{b^2}{4a}}\nonumber \end{equation}

For (2):

By using Fubini's Theorem, we first take the Laplace Transform with respect to $a$: \begin{equation} \mathscr{L}_{a \rightarrow s}\left[I(a,b)\right] = \int_0^\infty \mathscr{L}_{a \rightarrow s}\left[e^{-ax^2} \right] \cos(bx)\:dx = \int_0^\infty \frac{\cos(bx)}{s + x^2}\:dx \end{equation} We again employ Fubini's Theorem by taking the Laplace transform with respect to $b$: \begin{equation} \mathscr{L}_{b \rightarrow w}\left[\mathscr{L}_{a \rightarrow s}\left[I(a,b)\right]\right] = \int_0^\infty \frac{\mathscr{L}_{b \rightarrow w}\left[\cos(bx)\right]}{s + x^2}\:dx = \int_0^\infty \frac{w}{w^2 + x^2}\cdot \frac{1}{s + x^2}\:dx \end{equation} We proceed by forming a Partial Fraction Decomposition: \begin{align} &\mathscr{L}_{b \rightarrow w}\left[\mathscr{L}_{a \rightarrow s}\left[I(a,b)\right]\right] = \frac{w}{w^2 - s}\int_0^\infty\left(\frac{1}{s + x^2} - \frac{1}{w^2 + x^2} \right) \:dx \nonumber \\ &\quad= \frac{w}{w^2 - s}\left[ \frac{1}{\sqrt{s}}\arctan\left(\frac{x}{\sqrt{s}}\ \right) -\frac{1}{w}\arctan\left(\frac{x}{w} \right)\right]_0^\infty = \frac{w}{w^2 - s}\left[ \frac{1}{\sqrt{s}}\frac{\pi}{2} -\frac{1}{w}\frac{\pi}{2}\right] \nonumber \\ &= \frac{w}{w^2 - s}\left[\frac{w - \sqrt{s}}{w\sqrt{s}}\right]\frac{\pi}{2} = \frac{\pi}{2\sqrt{s}\left(w + \sqrt{s}\right)} \end{align} We now take the Inverse Laplace Transform with respect to $w$: \begin{equation} \mathscr{L}_{a \rightarrow s}\left[I(a,b)\right] = \mathscr{L}_{w \rightarrow b}^{-1}\left[ \frac{\pi}{2\sqrt{s}\left(w + \sqrt{s}\right)}\right] = \frac{\pi}{2\sqrt{s}}e^{-b\sqrt{s}} \end{equation} We now take the Inverse Laplace Transform with respect to $s$: \begin{equation} I(a,b) = \mathscr{L}_{s \rightarrow a}^{-1}\left[ \frac{\pi}{2\sqrt{s}}e^{-b\sqrt{s}} \right] = \frac{\pi}{2} \cdot \frac{1}{\sqrt{a\pi}}e^{-\frac{b^2}{4a}}= \frac{\sqrt{\pi}}{2\sqrt{a}}e^{-\frac{b^2}{4a}} \end{equation}

  • $\begingroup$ Excellent, complete, intutitive solution! +1 $\endgroup$ – clathratus Mar 4 at 3:41

For the first one, using @aleden's hint in the comments, substitute $$y = x^a, dy=ax^{a-1}dx = a\frac{y}{x}dx = a \frac{y}{y^{1/a}}dx \implies dx = \frac{1}{a}y^{1/a -1}dy$$ to reduce it to a Beta function:

$$ \int_0^\infty \frac{1}{1 +x^a} dx = \int_0^\infty \frac{1}{1+y}\frac{1}{a}y^{1/a -1}dy = \frac{1}{a}\int_0^\infty \frac{y^{1/a-1}}{1+y}dy = \frac{1}{a} B(1/a, 1 - 1/a) $$

Then use the well-known identity $$ B(m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} $$ to reduce the integral to $$ \frac{1}{a}\Gamma(1/a)\Gamma(1-1/a) $$ since $1/a + 1 -1/a = 1$ and $\Gamma(1) = 1$. Then use the Gamma function reflection formula $$ \Gamma(p)\Gamma(1-p) = \frac{\pi}{\sin p\pi} $$ to conclude that the integral is equal to $$ \frac{\pi/a}{\sin(\pi/a)} $$

For the second integral, using my hint in the comment, since the integrand is an even function of x, we can write the original integral as half the integral over the whole real line and then add ${i}$ times the similar integral with $\sin$ instead of $\cos$ - the latter is zero since the integrand is odd. We can then combine the two into a complex exponential, combine exponentials and complete the square in the exponent:

$$ \begin{align} \int \limits_{0}^{\infty} e^{-a x^2}\cos(b x) dx & = \frac{1}{2}\int \limits_{-\infty}^{\infty} e^{-a x^2}\cos(b x) dx\\ & = \frac{1}{2}\left (\int \limits_{-\infty}^{\infty} e^{-a x^2}\cos(b x) dx + i\int \limits_{-\infty}^{\infty} e^{-a x^2}\sin(b x) dx \right)\\ & = \frac{1}{2} \int \limits_{-\infty}^{\infty} e^{-a x^2}e^{ibx} dx\\ & = \frac{1}{2} \int \limits_{-\infty}^{\infty} e^{-a x^2 +ibx} dx\\ & = \frac{1}{2} e^{-b^2/4a} \int \limits_{-\infty}^{\infty} e^{-a (x^2 -i\frac{b}{a} x - b^2/4a^2)} dx\\ & = \frac{1}{2} e^{-b^2/4a} \int \limits_{-\infty}^{\infty} e^{-a (x -i \frac{b}{2a})^2} dx\\ & = \frac{1}{2} e^{-b^2/4a} \int \limits_{-\infty}^{\infty} e^{-ay^2} dy \end{align} $$ by making the substitution $y = x - i\frac{b}{2a}$. The Euler integral is calculated with the usual method and its value is $\sqrt{\frac{\pi}{a}}$, so the final value of the integral is

$$ \frac{1}{2} e^{-b^2/4a}\sqrt{\frac{\pi}{a}} $$


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