I stumbled upon this complicated looking integral while trying to solve an assignment for extra credit in Calculus.

$$\int^\infty_0 \frac{e^{-5x}- \cos x}{x} dx$$

I tried breaking it into two parts in order to deal with each separately. I tried the only substitution that made sense to me, but got nothing. Integrating by parts still yielded no significant progress. I thought about using Gamma and Beta functions, but I can't see how to build them up in this particular case

Last thing I tried before coming here for help was Taylor series, I thought that might help me get something more familiar...

Any ideas?


Let us recall the Mellin transform and its general form:

$$F(s) = \int^\infty_0 x^{s-1}f(x) dx$$

We might use this fact in this particular case, by trying to match our integral with $F(s)$. Hence, we might write:

$$\int^\infty_0 \frac{e^{-5x}- \cos x}{x} dx = \int^\infty_0 \frac{e^{-5x}}{x} dx - \int^\infty_0 \frac{\cos x}{x} dx = \int^\infty_0 x^{-1}e^{-5x} dx - \int^\infty_0 x^{-1}\cos x\ dx.$$

You see now how the general form of the Mellin transform now encapsulates our particular case, where $s \rightarrow0$. Therefore:

$$\int^\infty_0 x^{-1}e^{-5x} dx - \int^\infty_0 x^{-1}\cos x\ dx= \underset{s \to 0}{\lim} \{ \mathcal{Me^{-5x}}\ \}(s) - \{ \mathcal{M\cos x}\ \}(s) $$

This further yields our final answer:

$$\underset{s \to 0}{\lim} \{ \mathcal{Me^{-5x}}\ \}(s) - \{ \mathcal{M\cos x}\ \}(s) = \underset{s \to 0}{\lim}\ 5^{-s} \Gamma(s) - \Gamma(s)\cos(\frac{\pi\ s}{2}) = -\log(5).$$

For more reading on the Mellin transform, check this out.


Using $$ \frac1x=\int_0^\infty e^{-tx}\,\mathrm{d}t $$ we get $$ \begin{align} \int_0^\infty\frac{e^{-5x}-\cos(x)}{x}\mathrm{d}x &=\int_0^\infty\int_0^\infty\left(e^{-5x}-\cos(x)\right)e^{-tx}\,\mathrm{d}t\,\mathrm{d}x\\ &=\int_0^\infty\int_0^\infty\left(e^{-5x}-\cos(x)\right)e^{-tx}\,\mathrm{d}x\,\mathrm{d}t\\ &=\int_0^\infty\left(\frac1{5+t}-\frac{t}{1+t^2}\right)\,\mathrm{d}t\\ &=\lim_{a\to\infty}\int_0^a\left(\frac1{5+t}-\frac{t}{1+t^2}\right)\,\mathrm{d}t\\ &=\lim_{a\to\infty}\left[\log\left(\frac{5+a}5\right)-\frac12\log\left(1+a^2\right)\right]\\ &=\lim_{a\to\infty}\frac12\log\left(\frac{25+10a+a^2}{1+a^2}\right)-\log(5)\\[6pt] &=-\log(5) \end{align} $$


If you write your integral in the form: $$ I=\text{Re}\int_{0}^{+\infty}\frac{e^{-5x}-e^{-ix}}{x}\,dx $$ you may just apply some version of Frullani's theorem or of the residue theorem to get: $$ I = \color{red}{-\log 5}.$$

As an alternative, through the identity $$ \int_{0}^{+\infty}\frac{f(x)}{x}\,dx = \int_{0}^{+\infty}\mathcal{L}(f)(s)\,ds $$ that involves the Laplace transform, the problem boils down to computing: $$ \text{Re}\int_{0}^{+\infty}\left(\frac{1}{5+s}-\frac{1}{i+s}\right)\,ds=\int_{0}^{+\infty}\frac{1-5s}{(s+5)(s^2+1)}\,ds$$ that is straightforward.

  • $\begingroup$ How exactly did you combine those 2 fractions into one in that very last sentence? I couldn't get that somehow... $\endgroup$ – imranfat Sep 12 '15 at 1:53
  • $\begingroup$ @imranfat: see robjohn's answer. $\endgroup$ – Jack D'Aurizio Sep 12 '15 at 2:18
  • $\begingroup$ Yes I understand his steps, but his second fraction is different from yours. even if you multiply top and bottom by $s-i$ you do get that denom of $s^2+1$ but your num becomes $s-i$ and when you combine the 2 fractions, I just can't get $1-5s$ Rob's second fraction is different from yours. I am familiar with the Laplace transform so I dug into your method... $\endgroup$ – imranfat Sep 12 '15 at 2:33
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    $\begingroup$ @imranfat: $\frac1{s+i}=\frac{s-i}{s^2+1}$. Take the real part then you get $$\begin{align}\mathrm{Re}\left(\frac1{5+s}-\frac1{s+i}\right) &=\frac1{5+s}-\frac s{s^2+1}\\ &=\frac{1-5s}{(5+s)(s^2+1)}\end{align}$$ $\endgroup$ – robjohn Sep 12 '15 at 3:13
  • $\begingroup$ @robjohn Ah, knocked off the imaginary part and considering the real part. Yes, I understand that. The steps went a little too quick for me to grasp that... $\endgroup$ – imranfat Sep 12 '15 at 19:28

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